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Page 1
Number System
Basics of Number System
(1). Face Value : It is nothing but the
number itself about which it has been
asked.
Eg: In the number 23576, face value of 5
is 5 and face value of 7 is 7.
(2). Place Value : The place value of a
number depends on its position in the
number. Each position has a value 10
n
,
the places to its right.
Eg: In the number 23576, place value of 5
is 500 and place value of 3 is 3000.
Types of Numbers
(1). Natural Numbers (N) :
All positive counting numbers. (0 is not
included in it.)
Examples: 1, 2, 3, 4 … etc.
(2). Whole Numbers (W): All non-
negative numbers are all whole numbers.
Examples: 0, 1, 2, 3, 4… etc.
(3). Integer Numbers (I): All positive
numbers and negative numbers
including zero. Positive numbers are
called positive integers and negative
numbers are called negative integers.
I = ….. , -4, - 3, - 2, - 1, 0, 1, 2, 3, 4 …...
(4). Even Numbers : 2, 4, 6, 8, 10…..
[Divisible by 2 completely]
(5). Odd Numbers : 1, 3, 5, 7, 9, 11….. [Not
divisible by 2 completely]
(6). Rational Numbers : Numbers whose
exact value can be determined. Also a
number which can be written in the form
, where p and q are integers and q ? 0
??
??
is called a rational number. For example,
Examples : = 0.75, = 0.8 , ,
3
4
4
5
9
- 5
22
7
(7). Irrational Numbers : Numbers whose
exact value cannot be determined.
Example : = 3.142857142857 … p
(8). Prime number : A number which is
divisible by 1 and itself. 2 is only an even
prime number.
Example : 2, 3, 5, 7, 11, etc.
Note:-
Total prime no. between 1 - 50 15 ?
Total prime no. between 1 - 100 25 ?
Total prime no. between 1 - 500 95 ?
Total prime no. between 1 - 1000 168 ?
(9). Composite number : If we remove all
prime numbers from natural numbers
then whatever is left is called Composite
numbers.
Example : 4, 6, 8, 9, 10, 12 etc.
Note :- 1 is neither prime nor composite.
(10). Co - prime number : Two numbers
are called Co-prime numbers if their HCF
is 1.
Example : (2 and 3), (6 and 11).
Note : Two prime numbers are always
co-prime numbers to each other. Any two
consecutive integers are always co-prime
number to each other.
Factors
The factors of a number are the numbers
that divide it completely without leaving
any remainder.
Example : 24 can be completely divided
by 1, 2, 3, 4, 6, 8, 12 and 24, so these
numbers are factors of 24.
Prime factorisation of a number : When
a number is written in the form of
multiplication of its prime factors, it's
called prime factorisation.
Prime factorisation of 24.
24 2 2 2 3 or ? × × × 2
3
× 3
1
Number of factors : To ?nd the number
of factors we write the number in the
form of prime factors and then add +1 to
the exponent of prime factors and
multiply them.
For example: 24 = 2
3
× 3
1
Number of factors of 24 (3 + 1)(1 + 1) ?
= 4 2 = 8. ×
With the help of an example, we try to
?nd the sum of all factors of a number.
24 = , 2
3
× 3
1
Sum of all factors = ( + ) 2
0
+ 2
1
2
2
+ 2
3
×
( ) = 15 4 = 60. 3
0
+ 3
1
×
Number of even factors of a number : To
?nd the number of even factors of a
number, we add +1 to the exponents of
prime numbers except 2.
(Note : If a number doesn't have 2 as its
factor it will have 0 even factors)
Que . Find the number of even factors of
120.
Ans. 120 = 2
3
× 3
1
× 5
1
Number of even factors = 3 (1 + 1) × ×
(1 + 1) = 3 2 2 = 12 × ×
Note :- To ?nd the sum of even factors ,
we shall ignore , 2
0
Que. Find the sum of even factors of 120.
Sol:- Sum of even factors = ( + + ) 2
1
2
2
2
3
( )( ) = 14 4 6 = 336. 3
0
+ 3
1
5
0
+ 5
1
× ×
Number and Sum of odd factors of a
number : to ?nd the number and sum of
odd factors of a number, we have to
ignore the exponents of 2.
Que . Find the number of odd factors 120.
Sol:- 120 = 2
3
× 3
1
× 5
1
Required number = (1 + 1)(1 + 1) = 4
The exponent of 2 is completely ignored.
Sum of odd factors of 120 = ( )( 3
0
+ 3
1
) = 4 6 = 24 5
0
+ 5
1
×
Some Important Results of Factors:
1001 = 7 11 13 × ×
1001 abc = abcabc ×
1001 234 = 234234 ×
Que: Which of the following is a factor of
531531?
(a) 15 (b) 13 (c) 11 (d) both b and c
Sol:- 531531 = 1001 531 ×
= 7 11 13 531 So, both 11 and 13 × × ×
are factors of 531531.
111 = 37 3 , 1001 111 = 111111, × ×
When a single digit is written 6 times, 3,
7, 11, 13, and 37 are factors of it.
Que . Which of the following is a factor of
222222 ?
(a) 17 (b) 57 (c) 68 (d) 74
Sol :- 222222 = 2 111111 ×
= 2 37 × 3 × 7 × 11 × 13 ×
Clearly, 2 37 = 74 is one of the factors. ×
? If a, b and c are prime numbers, then
the number of prime factors of ??
??
× ??
??
×
is (x + y + z). ??
??
Recurring Decimal
Recurring decimals are referred to as
numbers that are uniformly repeated
after the decimal. Some rational
numbers produce recurring decimals
after converting them into decimal
numbers, but all irrational numbers
produce recurring decimals after
converting them into decimal form.
Examples :
(1) = 0.3333333 ….. = 0.
1
3
3
(2) 0. = = 1 9
9
9
(3) 0.53 = = 27
5327 - 53
9900
5274
9900
Page 2
Number System
Basics of Number System
(1). Face Value : It is nothing but the
number itself about which it has been
asked.
Eg: In the number 23576, face value of 5
is 5 and face value of 7 is 7.
(2). Place Value : The place value of a
number depends on its position in the
number. Each position has a value 10
n
,
the places to its right.
Eg: In the number 23576, place value of 5
is 500 and place value of 3 is 3000.
Types of Numbers
(1). Natural Numbers (N) :
All positive counting numbers. (0 is not
included in it.)
Examples: 1, 2, 3, 4 … etc.
(2). Whole Numbers (W): All non-
negative numbers are all whole numbers.
Examples: 0, 1, 2, 3, 4… etc.
(3). Integer Numbers (I): All positive
numbers and negative numbers
including zero. Positive numbers are
called positive integers and negative
numbers are called negative integers.
I = ….. , -4, - 3, - 2, - 1, 0, 1, 2, 3, 4 …...
(4). Even Numbers : 2, 4, 6, 8, 10…..
[Divisible by 2 completely]
(5). Odd Numbers : 1, 3, 5, 7, 9, 11….. [Not
divisible by 2 completely]
(6). Rational Numbers : Numbers whose
exact value can be determined. Also a
number which can be written in the form
, where p and q are integers and q ? 0
??
??
is called a rational number. For example,
Examples : = 0.75, = 0.8 , ,
3
4
4
5
9
- 5
22
7
(7). Irrational Numbers : Numbers whose
exact value cannot be determined.
Example : = 3.142857142857 … p
(8). Prime number : A number which is
divisible by 1 and itself. 2 is only an even
prime number.
Example : 2, 3, 5, 7, 11, etc.
Note:-
Total prime no. between 1 - 50 15 ?
Total prime no. between 1 - 100 25 ?
Total prime no. between 1 - 500 95 ?
Total prime no. between 1 - 1000 168 ?
(9). Composite number : If we remove all
prime numbers from natural numbers
then whatever is left is called Composite
numbers.
Example : 4, 6, 8, 9, 10, 12 etc.
Note :- 1 is neither prime nor composite.
(10). Co - prime number : Two numbers
are called Co-prime numbers if their HCF
is 1.
Example : (2 and 3), (6 and 11).
Note : Two prime numbers are always
co-prime numbers to each other. Any two
consecutive integers are always co-prime
number to each other.
Factors
The factors of a number are the numbers
that divide it completely without leaving
any remainder.
Example : 24 can be completely divided
by 1, 2, 3, 4, 6, 8, 12 and 24, so these
numbers are factors of 24.
Prime factorisation of a number : When
a number is written in the form of
multiplication of its prime factors, it's
called prime factorisation.
Prime factorisation of 24.
24 2 2 2 3 or ? × × × 2
3
× 3
1
Number of factors : To ?nd the number
of factors we write the number in the
form of prime factors and then add +1 to
the exponent of prime factors and
multiply them.
For example: 24 = 2
3
× 3
1
Number of factors of 24 (3 + 1)(1 + 1) ?
= 4 2 = 8. ×
With the help of an example, we try to
?nd the sum of all factors of a number.
24 = , 2
3
× 3
1
Sum of all factors = ( + ) 2
0
+ 2
1
2
2
+ 2
3
×
( ) = 15 4 = 60. 3
0
+ 3
1
×
Number of even factors of a number : To
?nd the number of even factors of a
number, we add +1 to the exponents of
prime numbers except 2.
(Note : If a number doesn't have 2 as its
factor it will have 0 even factors)
Que . Find the number of even factors of
120.
Ans. 120 = 2
3
× 3
1
× 5
1
Number of even factors = 3 (1 + 1) × ×
(1 + 1) = 3 2 2 = 12 × ×
Note :- To ?nd the sum of even factors ,
we shall ignore , 2
0
Que. Find the sum of even factors of 120.
Sol:- Sum of even factors = ( + + ) 2
1
2
2
2
3
( )( ) = 14 4 6 = 336. 3
0
+ 3
1
5
0
+ 5
1
× ×
Number and Sum of odd factors of a
number : to ?nd the number and sum of
odd factors of a number, we have to
ignore the exponents of 2.
Que . Find the number of odd factors 120.
Sol:- 120 = 2
3
× 3
1
× 5
1
Required number = (1 + 1)(1 + 1) = 4
The exponent of 2 is completely ignored.
Sum of odd factors of 120 = ( )( 3
0
+ 3
1
) = 4 6 = 24 5
0
+ 5
1
×
Some Important Results of Factors:
1001 = 7 11 13 × ×
1001 abc = abcabc ×
1001 234 = 234234 ×
Que: Which of the following is a factor of
531531?
(a) 15 (b) 13 (c) 11 (d) both b and c
Sol:- 531531 = 1001 531 ×
= 7 11 13 531 So, both 11 and 13 × × ×
are factors of 531531.
111 = 37 3 , 1001 111 = 111111, × ×
When a single digit is written 6 times, 3,
7, 11, 13, and 37 are factors of it.
Que . Which of the following is a factor of
222222 ?
(a) 17 (b) 57 (c) 68 (d) 74
Sol :- 222222 = 2 111111 ×
= 2 37 × 3 × 7 × 11 × 13 ×
Clearly, 2 37 = 74 is one of the factors. ×
? If a, b and c are prime numbers, then
the number of prime factors of ??
??
× ??
??
×
is (x + y + z). ??
??
Recurring Decimal
Recurring decimals are referred to as
numbers that are uniformly repeated
after the decimal. Some rational
numbers produce recurring decimals
after converting them into decimal
numbers, but all irrational numbers
produce recurring decimals after
converting them into decimal form.
Examples :
(1) = 0.3333333 ….. = 0.
1
3
3
(2) 0. = = 1 9
9
9
(3) 0.53 = = 27
5327 - 53
9900
5274
9900
(4) 2.53 = 2 + = 2 27
5327 - 53
9900
5274
9900
Divisibility Test
By 2:- When last digit is 0 or an even
number eg: 520, 588
By 3:- Sum of digits is divisible by 3
eg: 1971, 1974
By 4:- When last two digits are divisible
by 4 or, they are zeros eg: 1528, 1700
By 5 :- When last digit is 0 or 5
eg: 1725, 1790
By 6 :- When the number is divisible by 2
and 3 both. eg: 36, 72
By 7 : - Subtract twice the last digit from
the number formed by the remaining
digits. Like 651 divisible by 7
65 - (1 × 2) = 63. Since 63 is divisible by
7, so is 651.
By 8 :- When the last three digits are
divisible by 8. eg: 2256
By 9 :- When sum of digit is divisible by 9
eg: 9216
By 10 :- When the last digit is 0. eg:
452600
By 11:- When the difference between the
sum of odd and even place digits is
equal to 0 or multiple of 11 .
eg: 217382
Sum of odd place digits = 2 + 7 + 8 = 17
Sum of even place digits = 1 + 3 + 2 = 6
17 – 6 = 11, hence 217382 is divisible by
11.
By 13 : - If adding four times the last
digit to the number formed by the
remaining digits is divisible by 13, then
the number is divisible by 13. Like 1326
is divisible by 13
132 + (6 4) = 156. Repeat the same ×
process for 156 .
15 + (6 4) = 39.so 39 is divisible by 13 ×
BY 17 :- The divisibility rule of 17 states,
"If ?ve times the last digit is subtracted
from a number made up of the remaining
digits and the remainder is either 0 or a
multiple of 17, then the number is
divisible by 17".
Like 221: 22 - 1 × 5 = 17.
Prime Number Test
For ?nding whether any number is a
prime number or not, we need to ?nd the
nearest square root of given number,
then we need to ?nd out whether the
given number is divisible by any prime
number less than the obtained number or
not. If it is divisible then it is not a prime
number and if not divisible then it is a
prime number.
Example : Find whether 177 is a prime
number or not.
Soln : Nearest square root of 177 is 13.
Now we need to check whether 177 is
divisible by prime numbers less than 13.
On checking we ?nd that 177 is divisible
by 3. Hence, 177 is not a prime number.
Important Formulas
1. Sum of ?rst n natural number
s =
?? ( ?? + 1 )
2
2. Sum of ?rst n odd numbers = ??
2
3. Sum of ?rst n even numbers
= ?? ( ?? + 1 )
4. Sum of square of ?rst n natural
numbers =
?? ( ?? + 1 )( 2 ?? + 1 )
6
5. Sum of cubes of ?rst n natural
number = ( )
2
?? ( ?? + 1 )
2
6. is divisible by for ( ??
??
- ??
??
) ?? - ?? ( )
every natural number m.
7. is divisible by and ( ??
??
- ??
??
) ?? + ?? ( )
for even values of m. ?? - ?? ( )
8. is divisible by for ( ??
??
+ ??
??
) ?? + ?? ( )
odd values of m.
9. Number of prime factors of
is when a, ??
??
, ??
??
, ??
??
, ??
??
?? + ?? + ?? + ??
b, c, d are all prime numbers.
10. HCF of and = ( ??
??
- 1 ) ( ??
??
- 1 )
[ ] ( ??
?????? ( ?? , ?? )
- 1 )
Number of Zeros in an expression
We shall understand this concept with
the help of an example.
Let’s ?nd the number of zeros in the
following expression: 24 32 17 23 × × ×
19 = (2
3
× 3
1
) 2
5
× 17 × 23 × 19 × ×
Notice that a zero is made only when
there is a combination of 2 and 5. Since
there is no ‘5’ here there will be no zero in
the above expression.
Example:-
8 × 15 × 23 × 17 × 25 × 22 =
2
3
× ( 3
1
× 5
1
) × 23 × 17 × 5
2
× 2
1
× 11
In this expression there are 4 twos and 3
?ves. From this 3 pairs of can be 5×2
formed. Therefore, there will be 3 zeros
in the ?nal product.
Que . Find the number of zeros in the
value of:
2
2
× 5
4
× 4
6
× 10
8
× 6
10
× 15
12
× 8
14
. × 20
16
× 10
18
× 25
20
Sol:-
2
2
× 5
4
× 4
6
× 10
8
× 6
10
× 15
12
× 8
14
= × 20
16
× 10
18
× 25
20
2
2
× 5
4
× 2
12
× 2
8
× 5
8
× 2
10
× 3
10
× 3
12
× 5
12
× 2
42
× 2
32
× 5
16
× 2
18
× 5
18
× 5
40
Zeros are possible with a combination of
2 Here the number of 5’s are less so × 5
the number of zeros will be limited to the
number of 5’s.
In this expression number of ?ves are:
5
4
× 5
8
× 5
12
× 5
16
× 5
18
× 5
40
;
i.e. 4 + 8 + 12 + 16 + 18 + 40 = 98
The number of Zeros in n!
To ?nd the number of zeros in n!, we
divide “n” by 5 until we get a number less
than 5, and then we add all the quotients
so obtained.
Que. Find the number of zeros in 36! .
The number of zeros = 7 + 1 = 8.
Remainder Theorem
Que. What will be the remainder when
is divided by 12? 17 × 23
Ans :- We can express this as:
= 17 × 23 12 + 5 ( ) × 12 + 11 ( )
= 12 × 12 + 12 × 11 + 5 × 12 + 5 × 11
In the above expression we will ?nd that
remainder will depend on the last term
i.e. 5 × 11
Now, .( ) 7. ??????
5 × 11
12
=
So,
12 × 12 + 12 × 11 + 5 × 12 + 5 × 11
12
and remainder is same in both
5 × 11
12
cases which is 7.
Example:- Remainder when
is divided by 12? 1421 × 1423 × 1425
( ) ??????
1421 × 1423 × 1425
12
Page 3
Number System
Basics of Number System
(1). Face Value : It is nothing but the
number itself about which it has been
asked.
Eg: In the number 23576, face value of 5
is 5 and face value of 7 is 7.
(2). Place Value : The place value of a
number depends on its position in the
number. Each position has a value 10
n
,
the places to its right.
Eg: In the number 23576, place value of 5
is 500 and place value of 3 is 3000.
Types of Numbers
(1). Natural Numbers (N) :
All positive counting numbers. (0 is not
included in it.)
Examples: 1, 2, 3, 4 … etc.
(2). Whole Numbers (W): All non-
negative numbers are all whole numbers.
Examples: 0, 1, 2, 3, 4… etc.
(3). Integer Numbers (I): All positive
numbers and negative numbers
including zero. Positive numbers are
called positive integers and negative
numbers are called negative integers.
I = ….. , -4, - 3, - 2, - 1, 0, 1, 2, 3, 4 …...
(4). Even Numbers : 2, 4, 6, 8, 10…..
[Divisible by 2 completely]
(5). Odd Numbers : 1, 3, 5, 7, 9, 11….. [Not
divisible by 2 completely]
(6). Rational Numbers : Numbers whose
exact value can be determined. Also a
number which can be written in the form
, where p and q are integers and q ? 0
??
??
is called a rational number. For example,
Examples : = 0.75, = 0.8 , ,
3
4
4
5
9
- 5
22
7
(7). Irrational Numbers : Numbers whose
exact value cannot be determined.
Example : = 3.142857142857 … p
(8). Prime number : A number which is
divisible by 1 and itself. 2 is only an even
prime number.
Example : 2, 3, 5, 7, 11, etc.
Note:-
Total prime no. between 1 - 50 15 ?
Total prime no. between 1 - 100 25 ?
Total prime no. between 1 - 500 95 ?
Total prime no. between 1 - 1000 168 ?
(9). Composite number : If we remove all
prime numbers from natural numbers
then whatever is left is called Composite
numbers.
Example : 4, 6, 8, 9, 10, 12 etc.
Note :- 1 is neither prime nor composite.
(10). Co - prime number : Two numbers
are called Co-prime numbers if their HCF
is 1.
Example : (2 and 3), (6 and 11).
Note : Two prime numbers are always
co-prime numbers to each other. Any two
consecutive integers are always co-prime
number to each other.
Factors
The factors of a number are the numbers
that divide it completely without leaving
any remainder.
Example : 24 can be completely divided
by 1, 2, 3, 4, 6, 8, 12 and 24, so these
numbers are factors of 24.
Prime factorisation of a number : When
a number is written in the form of
multiplication of its prime factors, it's
called prime factorisation.
Prime factorisation of 24.
24 2 2 2 3 or ? × × × 2
3
× 3
1
Number of factors : To ?nd the number
of factors we write the number in the
form of prime factors and then add +1 to
the exponent of prime factors and
multiply them.
For example: 24 = 2
3
× 3
1
Number of factors of 24 (3 + 1)(1 + 1) ?
= 4 2 = 8. ×
With the help of an example, we try to
?nd the sum of all factors of a number.
24 = , 2
3
× 3
1
Sum of all factors = ( + ) 2
0
+ 2
1
2
2
+ 2
3
×
( ) = 15 4 = 60. 3
0
+ 3
1
×
Number of even factors of a number : To
?nd the number of even factors of a
number, we add +1 to the exponents of
prime numbers except 2.
(Note : If a number doesn't have 2 as its
factor it will have 0 even factors)
Que . Find the number of even factors of
120.
Ans. 120 = 2
3
× 3
1
× 5
1
Number of even factors = 3 (1 + 1) × ×
(1 + 1) = 3 2 2 = 12 × ×
Note :- To ?nd the sum of even factors ,
we shall ignore , 2
0
Que. Find the sum of even factors of 120.
Sol:- Sum of even factors = ( + + ) 2
1
2
2
2
3
( )( ) = 14 4 6 = 336. 3
0
+ 3
1
5
0
+ 5
1
× ×
Number and Sum of odd factors of a
number : to ?nd the number and sum of
odd factors of a number, we have to
ignore the exponents of 2.
Que . Find the number of odd factors 120.
Sol:- 120 = 2
3
× 3
1
× 5
1
Required number = (1 + 1)(1 + 1) = 4
The exponent of 2 is completely ignored.
Sum of odd factors of 120 = ( )( 3
0
+ 3
1
) = 4 6 = 24 5
0
+ 5
1
×
Some Important Results of Factors:
1001 = 7 11 13 × ×
1001 abc = abcabc ×
1001 234 = 234234 ×
Que: Which of the following is a factor of
531531?
(a) 15 (b) 13 (c) 11 (d) both b and c
Sol:- 531531 = 1001 531 ×
= 7 11 13 531 So, both 11 and 13 × × ×
are factors of 531531.
111 = 37 3 , 1001 111 = 111111, × ×
When a single digit is written 6 times, 3,
7, 11, 13, and 37 are factors of it.
Que . Which of the following is a factor of
222222 ?
(a) 17 (b) 57 (c) 68 (d) 74
Sol :- 222222 = 2 111111 ×
= 2 37 × 3 × 7 × 11 × 13 ×
Clearly, 2 37 = 74 is one of the factors. ×
? If a, b and c are prime numbers, then
the number of prime factors of ??
??
× ??
??
×
is (x + y + z). ??
??
Recurring Decimal
Recurring decimals are referred to as
numbers that are uniformly repeated
after the decimal. Some rational
numbers produce recurring decimals
after converting them into decimal
numbers, but all irrational numbers
produce recurring decimals after
converting them into decimal form.
Examples :
(1) = 0.3333333 ….. = 0.
1
3
3
(2) 0. = = 1 9
9
9
(3) 0.53 = = 27
5327 - 53
9900
5274
9900
(4) 2.53 = 2 + = 2 27
5327 - 53
9900
5274
9900
Divisibility Test
By 2:- When last digit is 0 or an even
number eg: 520, 588
By 3:- Sum of digits is divisible by 3
eg: 1971, 1974
By 4:- When last two digits are divisible
by 4 or, they are zeros eg: 1528, 1700
By 5 :- When last digit is 0 or 5
eg: 1725, 1790
By 6 :- When the number is divisible by 2
and 3 both. eg: 36, 72
By 7 : - Subtract twice the last digit from
the number formed by the remaining
digits. Like 651 divisible by 7
65 - (1 × 2) = 63. Since 63 is divisible by
7, so is 651.
By 8 :- When the last three digits are
divisible by 8. eg: 2256
By 9 :- When sum of digit is divisible by 9
eg: 9216
By 10 :- When the last digit is 0. eg:
452600
By 11:- When the difference between the
sum of odd and even place digits is
equal to 0 or multiple of 11 .
eg: 217382
Sum of odd place digits = 2 + 7 + 8 = 17
Sum of even place digits = 1 + 3 + 2 = 6
17 – 6 = 11, hence 217382 is divisible by
11.
By 13 : - If adding four times the last
digit to the number formed by the
remaining digits is divisible by 13, then
the number is divisible by 13. Like 1326
is divisible by 13
132 + (6 4) = 156. Repeat the same ×
process for 156 .
15 + (6 4) = 39.so 39 is divisible by 13 ×
BY 17 :- The divisibility rule of 17 states,
"If ?ve times the last digit is subtracted
from a number made up of the remaining
digits and the remainder is either 0 or a
multiple of 17, then the number is
divisible by 17".
Like 221: 22 - 1 × 5 = 17.
Prime Number Test
For ?nding whether any number is a
prime number or not, we need to ?nd the
nearest square root of given number,
then we need to ?nd out whether the
given number is divisible by any prime
number less than the obtained number or
not. If it is divisible then it is not a prime
number and if not divisible then it is a
prime number.
Example : Find whether 177 is a prime
number or not.
Soln : Nearest square root of 177 is 13.
Now we need to check whether 177 is
divisible by prime numbers less than 13.
On checking we ?nd that 177 is divisible
by 3. Hence, 177 is not a prime number.
Important Formulas
1. Sum of ?rst n natural number
s =
?? ( ?? + 1 )
2
2. Sum of ?rst n odd numbers = ??
2
3. Sum of ?rst n even numbers
= ?? ( ?? + 1 )
4. Sum of square of ?rst n natural
numbers =
?? ( ?? + 1 )( 2 ?? + 1 )
6
5. Sum of cubes of ?rst n natural
number = ( )
2
?? ( ?? + 1 )
2
6. is divisible by for ( ??
??
- ??
??
) ?? - ?? ( )
every natural number m.
7. is divisible by and ( ??
??
- ??
??
) ?? + ?? ( )
for even values of m. ?? - ?? ( )
8. is divisible by for ( ??
??
+ ??
??
) ?? + ?? ( )
odd values of m.
9. Number of prime factors of
is when a, ??
??
, ??
??
, ??
??
, ??
??
?? + ?? + ?? + ??
b, c, d are all prime numbers.
10. HCF of and = ( ??
??
- 1 ) ( ??
??
- 1 )
[ ] ( ??
?????? ( ?? , ?? )
- 1 )
Number of Zeros in an expression
We shall understand this concept with
the help of an example.
Let’s ?nd the number of zeros in the
following expression: 24 32 17 23 × × ×
19 = (2
3
× 3
1
) 2
5
× 17 × 23 × 19 × ×
Notice that a zero is made only when
there is a combination of 2 and 5. Since
there is no ‘5’ here there will be no zero in
the above expression.
Example:-
8 × 15 × 23 × 17 × 25 × 22 =
2
3
× ( 3
1
× 5
1
) × 23 × 17 × 5
2
× 2
1
× 11
In this expression there are 4 twos and 3
?ves. From this 3 pairs of can be 5×2
formed. Therefore, there will be 3 zeros
in the ?nal product.
Que . Find the number of zeros in the
value of:
2
2
× 5
4
× 4
6
× 10
8
× 6
10
× 15
12
× 8
14
. × 20
16
× 10
18
× 25
20
Sol:-
2
2
× 5
4
× 4
6
× 10
8
× 6
10
× 15
12
× 8
14
= × 20
16
× 10
18
× 25
20
2
2
× 5
4
× 2
12
× 2
8
× 5
8
× 2
10
× 3
10
× 3
12
× 5
12
× 2
42
× 2
32
× 5
16
× 2
18
× 5
18
× 5
40
Zeros are possible with a combination of
2 Here the number of 5’s are less so × 5
the number of zeros will be limited to the
number of 5’s.
In this expression number of ?ves are:
5
4
× 5
8
× 5
12
× 5
16
× 5
18
× 5
40
;
i.e. 4 + 8 + 12 + 16 + 18 + 40 = 98
The number of Zeros in n!
To ?nd the number of zeros in n!, we
divide “n” by 5 until we get a number less
than 5, and then we add all the quotients
so obtained.
Que. Find the number of zeros in 36! .
The number of zeros = 7 + 1 = 8.
Remainder Theorem
Que. What will be the remainder when
is divided by 12? 17 × 23
Ans :- We can express this as:
= 17 × 23 12 + 5 ( ) × 12 + 11 ( )
= 12 × 12 + 12 × 11 + 5 × 12 + 5 × 11
In the above expression we will ?nd that
remainder will depend on the last term
i.e. 5 × 11
Now, .( ) 7. ??????
5 × 11
12
=
So,
12 × 12 + 12 × 11 + 5 × 12 + 5 × 11
12
and remainder is same in both
5 × 11
12
cases which is 7.
Example:- Remainder when
is divided by 12? 1421 × 1423 × 1425
( ) ??????
1421 × 1423 × 1425
12
= ( ) ( ) ??????
5 × 7 × 9
12
= ??????
35 × 9
12
( ) = ??????
11 × 9
12
= 3
Negative Remainder
Taking a negative remainder will make
our calculation easier.
Examples
(i) rem ( ) = rem ( )
7 × 8
9
- 2 × - 1
9
= 2 1 = 2 - × -
(ii) rem ( ) = rem ( )
55 × 56
57
- 2 × - 1
57
= - 2 1 × - = 2
(iii) rem ( ) = rem ( )
7 × 10
9
- 2 × 1
9
= 2 1 - × = - 2 ???? , 7
Large Power Concepts
Look at the following examples:
(i) rem ( ) = ( )
28
12345
9
??????
27 + 1 ( )
12345
9
= rem ( ) =
1
12345
9
1
12345
= 1
(ii) rem ( )
26
12345
9
= ( ) ??????
27 - 1 ( )
12345
9
= rem ( ) = - = - 1 or 8
- 1
12345
9
1
12345
Application of Remainder
Theorem
Que . Find the last two digits of the
expression
? 22 × 31 × 44 × 27 × 37 × 43
Sol:- If we divide the above expression by
100, we will get the last two digits as
remainder.
( ) , ? ??????
22 × 31 × 44 × 27 × 37 × 43
100
dividing by 4 to make it simple
( ) = ??????
22 × 31 × 11 × 27 × 37 × 43
25
( ) = ??????
132 × 22 × 216
25
( ) ( ) = ??????
7 × 22 × 16
25
? ??????
4 × 16
25
( ) = ??????
14
25
= 14
Since we had divided by 4 initially now to
get the correct answer, we need to
multiply the remainder by 4.
So remainder will be which 14 × 4 = 56 ,
will also be the last two digits of the
expression.
Variety Questions
Q.1. A six-digit number 11p9q4 is
divisible by 24. Then the greatest
possible value for pq is:
SSC CGL Tier II (26/10/2023)
(a) 56 (b) 68 (c) 42 (d) 32
Q.2. The remainder of the term
when divided 9 + 9
2
+............. + 9
( 2 ?? + 1 )
by 6 is:
SSC CHSL 11/08/2023 (4th Shift)
(a) 1 (b) 4 (c) 2 (d) 3
Q.3. Two numbers, when divided by a
certain divisor, leave the remainder 57.
When sum of the two numbers is divided
by the same divisor, the remainder is 49.
The divisor is:
SSC CHSL 08/08/2023 (3rd Shift)
(a) 56 (b) 57 (c) 49 (d) 65
Q.4. In a division sum, the divisor is 11
times the quotient and 5 times the
remainder. If the remainder is 44, then
the dividend is:
SSC CHSL 07/08/2023 (4th Shift)
(a) 8888 (b) 4448 (c) 8444 (d) 4444
Q.5. What is the least value of x + y, if 10
digit number 780x533y24 is divisible by
88 ?
SSC CHSL 03/08/2023 (4th Shift)
(a) 4 (b) 3 (c) 1 (d) 2
Q.6. During a division, Pranjal mistakenly
took as the dividend a number that was
10% more than the original dividend. He
also mistakenly took as the divisor a
number that was 25% more than the
original divisor. If the correct quotient of
the original division problem was 25 and
the remainder was 0, what was the
quotient that Pranjal obtained, assuming
his calculations had no error ?
SSC CGL 17/07/2023 (4th shift)
(a) 21.75 (b) 21.25 (c) 28.75 (d) 22
Q.7. The six-digit number 7x1yyx is a
multiple of 33 for non-zero digits x and y.
Which of the following could be a
possible value of (x + y) ?
Matriculation Level 30/06/2023 (Shift - 4)
(a) 5 (b) 4 (c) 2 (d) 3
Q.8. A girl wants to plant trees in her
garden in rows in such a way that the
number of trees in each row to be the
same. There are 10 rows and the number
of trees in each row is 12, what is the
number of trees in each row, if there are
5 more rows ?
SSC MTS 17/05/2023 (Evening)
(a) 10 (b) 8 (c) 6 (d) 12
Q.9. What is the total number of factors
of the number 720 except 1 and the
number itself?
SSC CHSL 10/03/2023 (4th Shift)
(a) 29 (b) 27 (c) 32 (d) 28
Q.10. Which of the following is the
smallest among ( 14 ) , ( 12 ) , ( 16 )
1
3
1
2
1
6
&
( 25 ) ?
1
12
SSC CHSL 10/03/2023 (3rd Shift)
(a) ( 14 ) (b) ( 25 ) (c) ( 16 ) (d) ( 12 )
1
3
1
12
1
6
1
2
Q.11. Which of the following statements
is correct ?
I.The Value of 100
2
- 99
2
+ 98
2
- 97
2
+ 96
2
- 95
2
+ 94
2
- 93
2
+ ...... + 22
2
- 21
2
is 4840.
II. The value of ( ) ( ) ( ??
2
+
1
??
2
?? -
1
??
??
4
)( )( ) is K
16
- +
1
??
4
?? +
1
??
??
4
-
1
??
4
1
??
16
SSC CGL 13/12/2022 (3rd Shift)
(a) Neither I nor II (b) Both I and II
(c) Only II (d) Only I
Q.12. If the seven-digit number 52A6B7C
is divisible by 33, and A, B, C are primes,
then the maximum value of 2A+3B+C is:
SSC CGL 12/12/2022 (3rd Shift)
(a) 32 (b) 23 (c) 27 (d) 34
Q.13. If the 9 - digit number 83P93678Q
is divisible by 72, then what is the value
of ? ??
2
+ ??
2
+ 12
SSC CGL 05/12/2022 (4th Shift)
(a) 6 (b) 7 (c) 8 (d) 9
Q.14. In a test (+ 5) marks are given for
every correct answer and (-2) marks are
given for every incorrect answer. Jay
answered all the questions and scored
(-12) marks, though he got 4 correct
answers. How many of his answers were
INCORRECT ?
SSC CPO 11/11/2022 (Evening)
(a) 8 (b) 32 (c) 16 (d) 20
Q.15. What is the sum of all the common
terms between the given series S1 and
S2 ?
S1 = 2, 9, 16, ……., 632
S2 = 7, 11, 15, ……., 743
SSC CGL Tier II (08/08/2022)
(a) 6974 (b) 6750 (c) 7140 (d) 6860
Q.16. If the 7 - digit number x8942y4 is
divisible by 56, what is the value of ( + ??
2
y) for the largest value of y, where x and y
are natural numbers?
SSC CGL 11/04/2022 (Evening)
(a) 33 (b) 44 (c) 55 (d) 70
Q.17. Let p, q, r and s be positive natural
numbers having three exact factors
www.ssccglpinnacle.com Download Pinnacle Exam Preparation App 3
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Number System
Basics of Number System
(1). Face Value : It is nothing but the
number itself about which it has been
asked.
Eg: In the number 23576, face value of 5
is 5 and face value of 7 is 7.
(2). Place Value : The place value of a
number depends on its position in the
number. Each position has a value 10
n
,
the places to its right.
Eg: In the number 23576, place value of 5
is 500 and place value of 3 is 3000.
Types of Numbers
(1). Natural Numbers (N) :
All positive counting numbers. (0 is not
included in it.)
Examples: 1, 2, 3, 4 … etc.
(2). Whole Numbers (W): All non-
negative numbers are all whole numbers.
Examples: 0, 1, 2, 3, 4… etc.
(3). Integer Numbers (I): All positive
numbers and negative numbers
including zero. Positive numbers are
called positive integers and negative
numbers are called negative integers.
I = ….. , -4, - 3, - 2, - 1, 0, 1, 2, 3, 4 …...
(4). Even Numbers : 2, 4, 6, 8, 10…..
[Divisible by 2 completely]
(5). Odd Numbers : 1, 3, 5, 7, 9, 11….. [Not
divisible by 2 completely]
(6). Rational Numbers : Numbers whose
exact value can be determined. Also a
number which can be written in the form
, where p and q are integers and q ? 0
??
??
is called a rational number. For example,
Examples : = 0.75, = 0.8 , ,
3
4
4
5
9
- 5
22
7
(7). Irrational Numbers : Numbers whose
exact value cannot be determined.
Example : = 3.142857142857 … p
(8). Prime number : A number which is
divisible by 1 and itself. 2 is only an even
prime number.
Example : 2, 3, 5, 7, 11, etc.
Note:-
Total prime no. between 1 - 50 15 ?
Total prime no. between 1 - 100 25 ?
Total prime no. between 1 - 500 95 ?
Total prime no. between 1 - 1000 168 ?
(9). Composite number : If we remove all
prime numbers from natural numbers
then whatever is left is called Composite
numbers.
Example : 4, 6, 8, 9, 10, 12 etc.
Note :- 1 is neither prime nor composite.
(10). Co - prime number : Two numbers
are called Co-prime numbers if their HCF
is 1.
Example : (2 and 3), (6 and 11).
Note : Two prime numbers are always
co-prime numbers to each other. Any two
consecutive integers are always co-prime
number to each other.
Factors
The factors of a number are the numbers
that divide it completely without leaving
any remainder.
Example : 24 can be completely divided
by 1, 2, 3, 4, 6, 8, 12 and 24, so these
numbers are factors of 24.
Prime factorisation of a number : When
a number is written in the form of
multiplication of its prime factors, it's
called prime factorisation.
Prime factorisation of 24.
24 2 2 2 3 or ? × × × 2
3
× 3
1
Number of factors : To ?nd the number
of factors we write the number in the
form of prime factors and then add +1 to
the exponent of prime factors and
multiply them.
For example: 24 = 2
3
× 3
1
Number of factors of 24 (3 + 1)(1 + 1) ?
= 4 2 = 8. ×
With the help of an example, we try to
?nd the sum of all factors of a number.
24 = , 2
3
× 3
1
Sum of all factors = ( + ) 2
0
+ 2
1
2
2
+ 2
3
×
( ) = 15 4 = 60. 3
0
+ 3
1
×
Number of even factors of a number : To
?nd the number of even factors of a
number, we add +1 to the exponents of
prime numbers except 2.
(Note : If a number doesn't have 2 as its
factor it will have 0 even factors)
Que . Find the number of even factors of
120.
Ans. 120 = 2
3
× 3
1
× 5
1
Number of even factors = 3 (1 + 1) × ×
(1 + 1) = 3 2 2 = 12 × ×
Note :- To ?nd the sum of even factors ,
we shall ignore , 2
0
Que. Find the sum of even factors of 120.
Sol:- Sum of even factors = ( + + ) 2
1
2
2
2
3
( )( ) = 14 4 6 = 336. 3
0
+ 3
1
5
0
+ 5
1
× ×
Number and Sum of odd factors of a
number : to ?nd the number and sum of
odd factors of a number, we have to
ignore the exponents of 2.
Que . Find the number of odd factors 120.
Sol:- 120 = 2
3
× 3
1
× 5
1
Required number = (1 + 1)(1 + 1) = 4
The exponent of 2 is completely ignored.
Sum of odd factors of 120 = ( )( 3
0
+ 3
1
) = 4 6 = 24 5
0
+ 5
1
×
Some Important Results of Factors:
1001 = 7 11 13 × ×
1001 abc = abcabc ×
1001 234 = 234234 ×
Que: Which of the following is a factor of
531531?
(a) 15 (b) 13 (c) 11 (d) both b and c
Sol:- 531531 = 1001 531 ×
= 7 11 13 531 So, both 11 and 13 × × ×
are factors of 531531.
111 = 37 3 , 1001 111 = 111111, × ×
When a single digit is written 6 times, 3,
7, 11, 13, and 37 are factors of it.
Que . Which of the following is a factor of
222222 ?
(a) 17 (b) 57 (c) 68 (d) 74
Sol :- 222222 = 2 111111 ×
= 2 37 × 3 × 7 × 11 × 13 ×
Clearly, 2 37 = 74 is one of the factors. ×
? If a, b and c are prime numbers, then
the number of prime factors of ??
??
× ??
??
×
is (x + y + z). ??
??
Recurring Decimal
Recurring decimals are referred to as
numbers that are uniformly repeated
after the decimal. Some rational
numbers produce recurring decimals
after converting them into decimal
numbers, but all irrational numbers
produce recurring decimals after
converting them into decimal form.
Examples :
(1) = 0.3333333 ….. = 0.
1
3
3
(2) 0. = = 1 9
9
9
(3) 0.53 = = 27
5327 - 53
9900
5274
9900
(4) 2.53 = 2 + = 2 27
5327 - 53
9900
5274
9900
Divisibility Test
By 2:- When last digit is 0 or an even
number eg: 520, 588
By 3:- Sum of digits is divisible by 3
eg: 1971, 1974
By 4:- When last two digits are divisible
by 4 or, they are zeros eg: 1528, 1700
By 5 :- When last digit is 0 or 5
eg: 1725, 1790
By 6 :- When the number is divisible by 2
and 3 both. eg: 36, 72
By 7 : - Subtract twice the last digit from
the number formed by the remaining
digits. Like 651 divisible by 7
65 - (1 × 2) = 63. Since 63 is divisible by
7, so is 651.
By 8 :- When the last three digits are
divisible by 8. eg: 2256
By 9 :- When sum of digit is divisible by 9
eg: 9216
By 10 :- When the last digit is 0. eg:
452600
By 11:- When the difference between the
sum of odd and even place digits is
equal to 0 or multiple of 11 .
eg: 217382
Sum of odd place digits = 2 + 7 + 8 = 17
Sum of even place digits = 1 + 3 + 2 = 6
17 – 6 = 11, hence 217382 is divisible by
11.
By 13 : - If adding four times the last
digit to the number formed by the
remaining digits is divisible by 13, then
the number is divisible by 13. Like 1326
is divisible by 13
132 + (6 4) = 156. Repeat the same ×
process for 156 .
15 + (6 4) = 39.so 39 is divisible by 13 ×
BY 17 :- The divisibility rule of 17 states,
"If ?ve times the last digit is subtracted
from a number made up of the remaining
digits and the remainder is either 0 or a
multiple of 17, then the number is
divisible by 17".
Like 221: 22 - 1 × 5 = 17.
Prime Number Test
For ?nding whether any number is a
prime number or not, we need to ?nd the
nearest square root of given number,
then we need to ?nd out whether the
given number is divisible by any prime
number less than the obtained number or
not. If it is divisible then it is not a prime
number and if not divisible then it is a
prime number.
Example : Find whether 177 is a prime
number or not.
Soln : Nearest square root of 177 is 13.
Now we need to check whether 177 is
divisible by prime numbers less than 13.
On checking we ?nd that 177 is divisible
by 3. Hence, 177 is not a prime number.
Important Formulas
1. Sum of ?rst n natural number
s =
?? ( ?? + 1 )
2
2. Sum of ?rst n odd numbers = ??
2
3. Sum of ?rst n even numbers
= ?? ( ?? + 1 )
4. Sum of square of ?rst n natural
numbers =
?? ( ?? + 1 )( 2 ?? + 1 )
6
5. Sum of cubes of ?rst n natural
number = ( )
2
?? ( ?? + 1 )
2
6. is divisible by for ( ??
??
- ??
??
) ?? - ?? ( )
every natural number m.
7. is divisible by and ( ??
??
- ??
??
) ?? + ?? ( )
for even values of m. ?? - ?? ( )
8. is divisible by for ( ??
??
+ ??
??
) ?? + ?? ( )
odd values of m.
9. Number of prime factors of
is when a, ??
??
, ??
??
, ??
??
, ??
??
?? + ?? + ?? + ??
b, c, d are all prime numbers.
10. HCF of and = ( ??
??
- 1 ) ( ??
??
- 1 )
[ ] ( ??
?????? ( ?? , ?? )
- 1 )
Number of Zeros in an expression
We shall understand this concept with
the help of an example.
Let’s ?nd the number of zeros in the
following expression: 24 32 17 23 × × ×
19 = (2
3
× 3
1
) 2
5
× 17 × 23 × 19 × ×
Notice that a zero is made only when
there is a combination of 2 and 5. Since
there is no ‘5’ here there will be no zero in
the above expression.
Example:-
8 × 15 × 23 × 17 × 25 × 22 =
2
3
× ( 3
1
× 5
1
) × 23 × 17 × 5
2
× 2
1
× 11
In this expression there are 4 twos and 3
?ves. From this 3 pairs of can be 5×2
formed. Therefore, there will be 3 zeros
in the ?nal product.
Que . Find the number of zeros in the
value of:
2
2
× 5
4
× 4
6
× 10
8
× 6
10
× 15
12
× 8
14
. × 20
16
× 10
18
× 25
20
Sol:-
2
2
× 5
4
× 4
6
× 10
8
× 6
10
× 15
12
× 8
14
= × 20
16
× 10
18
× 25
20
2
2
× 5
4
× 2
12
× 2
8
× 5
8
× 2
10
× 3
10
× 3
12
× 5
12
× 2
42
× 2
32
× 5
16
× 2
18
× 5
18
× 5
40
Zeros are possible with a combination of
2 Here the number of 5’s are less so × 5
the number of zeros will be limited to the
number of 5’s.
In this expression number of ?ves are:
5
4
× 5
8
× 5
12
× 5
16
× 5
18
× 5
40
;
i.e. 4 + 8 + 12 + 16 + 18 + 40 = 98
The number of Zeros in n!
To ?nd the number of zeros in n!, we
divide “n” by 5 until we get a number less
than 5, and then we add all the quotients
so obtained.
Que. Find the number of zeros in 36! .
The number of zeros = 7 + 1 = 8.
Remainder Theorem
Que. What will be the remainder when
is divided by 12? 17 × 23
Ans :- We can express this as:
= 17 × 23 12 + 5 ( ) × 12 + 11 ( )
= 12 × 12 + 12 × 11 + 5 × 12 + 5 × 11
In the above expression we will ?nd that
remainder will depend on the last term
i.e. 5 × 11
Now, .( ) 7. ??????
5 × 11
12
=
So,
12 × 12 + 12 × 11 + 5 × 12 + 5 × 11
12
and remainder is same in both
5 × 11
12
cases which is 7.
Example:- Remainder when
is divided by 12? 1421 × 1423 × 1425
( ) ??????
1421 × 1423 × 1425
12
= ( ) ( ) ??????
5 × 7 × 9
12
= ??????
35 × 9
12
( ) = ??????
11 × 9
12
= 3
Negative Remainder
Taking a negative remainder will make
our calculation easier.
Examples
(i) rem ( ) = rem ( )
7 × 8
9
- 2 × - 1
9
= 2 1 = 2 - × -
(ii) rem ( ) = rem ( )
55 × 56
57
- 2 × - 1
57
= - 2 1 × - = 2
(iii) rem ( ) = rem ( )
7 × 10
9
- 2 × 1
9
= 2 1 - × = - 2 ???? , 7
Large Power Concepts
Look at the following examples:
(i) rem ( ) = ( )
28
12345
9
??????
27 + 1 ( )
12345
9
= rem ( ) =
1
12345
9
1
12345
= 1
(ii) rem ( )
26
12345
9
= ( ) ??????
27 - 1 ( )
12345
9
= rem ( ) = - = - 1 or 8
- 1
12345
9
1
12345
Application of Remainder
Theorem
Que . Find the last two digits of the
expression
? 22 × 31 × 44 × 27 × 37 × 43
Sol:- If we divide the above expression by
100, we will get the last two digits as
remainder.
( ) , ? ??????
22 × 31 × 44 × 27 × 37 × 43
100
dividing by 4 to make it simple
( ) = ??????
22 × 31 × 11 × 27 × 37 × 43
25
( ) = ??????
132 × 22 × 216
25
( ) ( ) = ??????
7 × 22 × 16
25
? ??????
4 × 16
25
( ) = ??????
14
25
= 14
Since we had divided by 4 initially now to
get the correct answer, we need to
multiply the remainder by 4.
So remainder will be which 14 × 4 = 56 ,
will also be the last two digits of the
expression.
Variety Questions
Q.1. A six-digit number 11p9q4 is
divisible by 24. Then the greatest
possible value for pq is:
SSC CGL Tier II (26/10/2023)
(a) 56 (b) 68 (c) 42 (d) 32
Q.2. The remainder of the term
when divided 9 + 9
2
+............. + 9
( 2 ?? + 1 )
by 6 is:
SSC CHSL 11/08/2023 (4th Shift)
(a) 1 (b) 4 (c) 2 (d) 3
Q.3. Two numbers, when divided by a
certain divisor, leave the remainder 57.
When sum of the two numbers is divided
by the same divisor, the remainder is 49.
The divisor is:
SSC CHSL 08/08/2023 (3rd Shift)
(a) 56 (b) 57 (c) 49 (d) 65
Q.4. In a division sum, the divisor is 11
times the quotient and 5 times the
remainder. If the remainder is 44, then
the dividend is:
SSC CHSL 07/08/2023 (4th Shift)
(a) 8888 (b) 4448 (c) 8444 (d) 4444
Q.5. What is the least value of x + y, if 10
digit number 780x533y24 is divisible by
88 ?
SSC CHSL 03/08/2023 (4th Shift)
(a) 4 (b) 3 (c) 1 (d) 2
Q.6. During a division, Pranjal mistakenly
took as the dividend a number that was
10% more than the original dividend. He
also mistakenly took as the divisor a
number that was 25% more than the
original divisor. If the correct quotient of
the original division problem was 25 and
the remainder was 0, what was the
quotient that Pranjal obtained, assuming
his calculations had no error ?
SSC CGL 17/07/2023 (4th shift)
(a) 21.75 (b) 21.25 (c) 28.75 (d) 22
Q.7. The six-digit number 7x1yyx is a
multiple of 33 for non-zero digits x and y.
Which of the following could be a
possible value of (x + y) ?
Matriculation Level 30/06/2023 (Shift - 4)
(a) 5 (b) 4 (c) 2 (d) 3
Q.8. A girl wants to plant trees in her
garden in rows in such a way that the
number of trees in each row to be the
same. There are 10 rows and the number
of trees in each row is 12, what is the
number of trees in each row, if there are
5 more rows ?
SSC MTS 17/05/2023 (Evening)
(a) 10 (b) 8 (c) 6 (d) 12
Q.9. What is the total number of factors
of the number 720 except 1 and the
number itself?
SSC CHSL 10/03/2023 (4th Shift)
(a) 29 (b) 27 (c) 32 (d) 28
Q.10. Which of the following is the
smallest among ( 14 ) , ( 12 ) , ( 16 )
1
3
1
2
1
6
&
( 25 ) ?
1
12
SSC CHSL 10/03/2023 (3rd Shift)
(a) ( 14 ) (b) ( 25 ) (c) ( 16 ) (d) ( 12 )
1
3
1
12
1
6
1
2
Q.11. Which of the following statements
is correct ?
I.The Value of 100
2
- 99
2
+ 98
2
- 97
2
+ 96
2
- 95
2
+ 94
2
- 93
2
+ ...... + 22
2
- 21
2
is 4840.
II. The value of ( ) ( ) ( ??
2
+
1
??
2
?? -
1
??
??
4
)( )( ) is K
16
- +
1
??
4
?? +
1
??
??
4
-
1
??
4
1
??
16
SSC CGL 13/12/2022 (3rd Shift)
(a) Neither I nor II (b) Both I and II
(c) Only II (d) Only I
Q.12. If the seven-digit number 52A6B7C
is divisible by 33, and A, B, C are primes,
then the maximum value of 2A+3B+C is:
SSC CGL 12/12/2022 (3rd Shift)
(a) 32 (b) 23 (c) 27 (d) 34
Q.13. If the 9 - digit number 83P93678Q
is divisible by 72, then what is the value
of ? ??
2
+ ??
2
+ 12
SSC CGL 05/12/2022 (4th Shift)
(a) 6 (b) 7 (c) 8 (d) 9
Q.14. In a test (+ 5) marks are given for
every correct answer and (-2) marks are
given for every incorrect answer. Jay
answered all the questions and scored
(-12) marks, though he got 4 correct
answers. How many of his answers were
INCORRECT ?
SSC CPO 11/11/2022 (Evening)
(a) 8 (b) 32 (c) 16 (d) 20
Q.15. What is the sum of all the common
terms between the given series S1 and
S2 ?
S1 = 2, 9, 16, ……., 632
S2 = 7, 11, 15, ……., 743
SSC CGL Tier II (08/08/2022)
(a) 6974 (b) 6750 (c) 7140 (d) 6860
Q.16. If the 7 - digit number x8942y4 is
divisible by 56, what is the value of ( + ??
2
y) for the largest value of y, where x and y
are natural numbers?
SSC CGL 11/04/2022 (Evening)
(a) 33 (b) 44 (c) 55 (d) 70
Q.17. Let p, q, r and s be positive natural
numbers having three exact factors
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Pinnacle Day: 1st - 5th Number System
including 1 and the number itself If q > p
and both are two-digit numbers, and r > s
and both are one-digit numbers, then the
value of the expression is:
?? - ?? - 1
?? - ??
SSC CGL Tier II (03/02/2022)
(a) - s - 1 (b) s - 1 (c) 1 - s (d) s + 1
Q.18. Three fractions x, y and z are such
that x > y > z. When the smallest of them
is divided by the greatest, the result is
which exceeds y by 0.0625. If x + y +
9
16
z = 2 , then what is the value of x + z ?
3
12
SSC CGL Tier II (29/01/2022)
(a) (b) (c) (d)
5
4
1
4
7
4
3
4
Q.19. The six-digit number 537xy5 is
divisible by 125. How many such six -
digit numbers are there?
SSC CHSL 19/04/2021 (Morning)
(a) 4 (b) 2 (c) 3 (d) 5
Q.20. How many numbers between 400
and 700 are divisible by 5, 6 and 7 ?
SSC CPO 24/11/2020 (Evening)
(a) 2 (b) 5 (c) 10 (d) 20
Q.21. Find the number of prime factors in
the product . ( 30 )
5
× ( 24 )
5
SSC CGL Tier II (18/11/2020)
(a) 45 (b) 35 (c) 10 (d) 30
Q.22. Let ab, a b, is a 2-digit prime ?
number such that ba is also a prime
number. The sum of all such number is:
SSC CGL Tier II (16/11/2020)
(a) 374 (b) 418 (c) 407 (d) 396
Q.23. Given that 2
20
+ 1 is completely
divisible by a whole number. Which of the
following is completely divisible by the
same number ?
SSC CHSL 16/10/2020 (Afternoon)
(a) 2
15
+ 1 (b) 5 2
30
×
(c) 2
90
+ 1 (d) 2
60
+ 1
Q.24. Which of the following numbers
will completely divide 7
81
+ 7
82
+ 7
83
?
SSC CHSL 17/03/2020 (Morning)
(a) 399 (b) 389 (c) 387 (d) 397
Q.25. When a positive integer is divided
by d, the remainder is 15. When ten times
of the same number is divided by d, the
remainder is 6. The least possible value
of d is:
SSC CGL 05/03/2020 (Afternoon)
(a) 9 (b) 12 (c) 16 (d) 18
Q.26. When 200 is divided by a positive
integer x, the remainder is 8. How many
values of x are there?
SSC CGL 03/03/2020 (Afternoon)
(a) 7 (b) 5 (c) 8 (d) 6
Q.27. The number 1563241234351 is :
SSC CPO 13/12/2019 (Evening)
(a) divisible by both 3 and 11
(b) divisible by 11 but not by 3
(c) neither divisible by 3 nor by 11
(d) divisible by 3 but not by 11
Q.28. How many natural numbers less
than 1000 are divisible by 5 or 7 but NOT
by 35 ?
SSC CPO 11/12/2019 (Morning)
(a) 285 (b) 313 (c) 341 (d) 243
Q.29. If r is the remainder when each of
4749, 5601 and 7092 is divided by the
greatest possible number d ( 1), then >
the value of (d + r) will be:
SSC CPO 11/12/2019 (Morning)
(a) 276 (b) 271 (c) 298 (d) 282
Q.30. Let x be the least 4-digit number
which when divided by 2, 3, 4, 5, 6 and 7
leaves a remainder of 1 in each case. If x
lies between 2800 and 3000, then what is
the sum of digits of x ?
SSC CPO 09/12/2019 (Evening)
(a) 15 (b) 16 (c) 12 (d) 13
Q.31. If the six digit number 479xyz is
exactly divisible by 7, 11 and 13 , then {( y
+ z ) ÷ x} is equal to :
SSC CPO 09/12/2019 (Morning)
(a) (b) 4 (c) (d)
11
9
13
7
7
13
Q.32. Which among the following is the
smallest?
SSC CPO 09/12/19 (Morning)
(a) v401 - v399 (b) v101 - v99
(c) v301 - v299 (d) v201 - v199
Q.33. If x is the remainder when 3
61284
is
divided by 5 and y is the remainder when
4
96
is divided by 6, then what is the value
of (2x - y) ?
SSC CGL Tier II (13/09/2019)
(a) - 4 (b) 4 (c) -2 (d) 2
Q.34. In ?nding the HCF of two numbers
by division method, the last divisor is 17
and the quotients are 1, 11 and 2,
respectively. What is the sum of the two
numbers ?
SSC CGL Tier II (13/09/2019)
(a) 833 (b) 867 (c) 816 (d) 901
Q.35. Two positive numbers differ by
2001. When the larger number is divided
by the smaller number, the quotient is 9
and the remainder is 41. The sum of the
digits of the larger number is :
SSC CGL Tier II (13/09/2019)
(a) 15 (b) 11 (c) 10 (d) 14
Q.36. When a two-digit number is
multiplied by the sum of its digits, the
product is 424. When the number
obtained by interchanging its digits is
multiplied by the sum of the digits, the
result is 280. The sum of the digits of the
given number is :
SSC CGL Tier II (12/09/2019)
(a) 6 (b) 9 (c) 8 (d) 7
Q.37. Let x be the least number which
when divided by 15,18,20 and 27, the
remainder in each case is 10 and x is a
multiple of 31. What least number should
be added to x to make it a perfect square ?
SSC CGL Tier II (12/09/2019)
(a) 39 (b) 37 (c) 43 (d) 36
Q.38. The number of factors of 3600 is :
SSC CGL Tier II (12/09/2019)
(a) 45 (b) 44 (c) 43 (d) 42
Q.39. When 12, 16, 18, 20 and 25 divide
the least number x, the remainder in each
case is 4 but x is divisible by 7. What is
the digit at the thousands’ place in x ?
SSC CGL Tier II (11/09/2019)
(a) 5 (b) 8 (c) 4 (d) 3
Q.40. One of the factors of ( ), 8
2 ??
+ 5
2 ??
where k is an odd number, is :
SSC CGL Tier II (11/09/2019)
(a) 86 (b) 88 (c) 84 (d) 89
Q.41. Let x = ( 633 )
24
- ( 277 )
38
+
What is the unit digit of x ? ( 266 )
54
SSC CGL Tier II (11/09/2019)
(a) 7 (b) 6 (c) 4 (d) 8
Q.42. The sum of the digits of a two-digit
number is of the number. The unit
1
7
digit is 4 less than the tens digit. If the
number obtained on reversing its digit is
divided by 7, the remainder will be :
SSC CGL Tier II (11/09/2019)
(a) 4 (b) 5 (c) 1 (d) 6
Q.43. When 6892, 7105 and 7531 are
divided by the greatest number x, then
the remainder in each case is y. What is
the value of (x - y) ?
SSC MTS 22/08/2019 (Afternoon)
(a) 123 (b) 137 (c) 147 (d) 113
Q.44. Let x be the greatest number which
when divides 6475, 4984 and 4132, the
remainder in each case is the same.
What is the sum of digits of x?
SSC MTS 22/08/2019 (Morning)
(a) 4 (b) 7 (c) 5 (d) 6
Q.45. When an integer n is divided by 8,
the remainder is 3. What will be the
remainder if 6n - 1 is divided by 8 ?
SSC CGL 13/06/2019 (Evening)
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Page 5
Number System
Basics of Number System
(1). Face Value : It is nothing but the
number itself about which it has been
asked.
Eg: In the number 23576, face value of 5
is 5 and face value of 7 is 7.
(2). Place Value : The place value of a
number depends on its position in the
number. Each position has a value 10
n
,
the places to its right.
Eg: In the number 23576, place value of 5
is 500 and place value of 3 is 3000.
Types of Numbers
(1). Natural Numbers (N) :
All positive counting numbers. (0 is not
included in it.)
Examples: 1, 2, 3, 4 … etc.
(2). Whole Numbers (W): All non-
negative numbers are all whole numbers.
Examples: 0, 1, 2, 3, 4… etc.
(3). Integer Numbers (I): All positive
numbers and negative numbers
including zero. Positive numbers are
called positive integers and negative
numbers are called negative integers.
I = ….. , -4, - 3, - 2, - 1, 0, 1, 2, 3, 4 …...
(4). Even Numbers : 2, 4, 6, 8, 10…..
[Divisible by 2 completely]
(5). Odd Numbers : 1, 3, 5, 7, 9, 11….. [Not
divisible by 2 completely]
(6). Rational Numbers : Numbers whose
exact value can be determined. Also a
number which can be written in the form
, where p and q are integers and q ? 0
??
??
is called a rational number. For example,
Examples : = 0.75, = 0.8 , ,
3
4
4
5
9
- 5
22
7
(7). Irrational Numbers : Numbers whose
exact value cannot be determined.
Example : = 3.142857142857 … p
(8). Prime number : A number which is
divisible by 1 and itself. 2 is only an even
prime number.
Example : 2, 3, 5, 7, 11, etc.
Note:-
Total prime no. between 1 - 50 15 ?
Total prime no. between 1 - 100 25 ?
Total prime no. between 1 - 500 95 ?
Total prime no. between 1 - 1000 168 ?
(9). Composite number : If we remove all
prime numbers from natural numbers
then whatever is left is called Composite
numbers.
Example : 4, 6, 8, 9, 10, 12 etc.
Note :- 1 is neither prime nor composite.
(10). Co - prime number : Two numbers
are called Co-prime numbers if their HCF
is 1.
Example : (2 and 3), (6 and 11).
Note : Two prime numbers are always
co-prime numbers to each other. Any two
consecutive integers are always co-prime
number to each other.
Factors
The factors of a number are the numbers
that divide it completely without leaving
any remainder.
Example : 24 can be completely divided
by 1, 2, 3, 4, 6, 8, 12 and 24, so these
numbers are factors of 24.
Prime factorisation of a number : When
a number is written in the form of
multiplication of its prime factors, it's
called prime factorisation.
Prime factorisation of 24.
24 2 2 2 3 or ? × × × 2
3
× 3
1
Number of factors : To ?nd the number
of factors we write the number in the
form of prime factors and then add +1 to
the exponent of prime factors and
multiply them.
For example: 24 = 2
3
× 3
1
Number of factors of 24 (3 + 1)(1 + 1) ?
= 4 2 = 8. ×
With the help of an example, we try to
?nd the sum of all factors of a number.
24 = , 2
3
× 3
1
Sum of all factors = ( + ) 2
0
+ 2
1
2
2
+ 2
3
×
( ) = 15 4 = 60. 3
0
+ 3
1
×
Number of even factors of a number : To
?nd the number of even factors of a
number, we add +1 to the exponents of
prime numbers except 2.
(Note : If a number doesn't have 2 as its
factor it will have 0 even factors)
Que . Find the number of even factors of
120.
Ans. 120 = 2
3
× 3
1
× 5
1
Number of even factors = 3 (1 + 1) × ×
(1 + 1) = 3 2 2 = 12 × ×
Note :- To ?nd the sum of even factors ,
we shall ignore , 2
0
Que. Find the sum of even factors of 120.
Sol:- Sum of even factors = ( + + ) 2
1
2
2
2
3
( )( ) = 14 4 6 = 336. 3
0
+ 3
1
5
0
+ 5
1
× ×
Number and Sum of odd factors of a
number : to ?nd the number and sum of
odd factors of a number, we have to
ignore the exponents of 2.
Que . Find the number of odd factors 120.
Sol:- 120 = 2
3
× 3
1
× 5
1
Required number = (1 + 1)(1 + 1) = 4
The exponent of 2 is completely ignored.
Sum of odd factors of 120 = ( )( 3
0
+ 3
1
) = 4 6 = 24 5
0
+ 5
1
×
Some Important Results of Factors:
1001 = 7 11 13 × ×
1001 abc = abcabc ×
1001 234 = 234234 ×
Que: Which of the following is a factor of
531531?
(a) 15 (b) 13 (c) 11 (d) both b and c
Sol:- 531531 = 1001 531 ×
= 7 11 13 531 So, both 11 and 13 × × ×
are factors of 531531.
111 = 37 3 , 1001 111 = 111111, × ×
When a single digit is written 6 times, 3,
7, 11, 13, and 37 are factors of it.
Que . Which of the following is a factor of
222222 ?
(a) 17 (b) 57 (c) 68 (d) 74
Sol :- 222222 = 2 111111 ×
= 2 37 × 3 × 7 × 11 × 13 ×
Clearly, 2 37 = 74 is one of the factors. ×
? If a, b and c are prime numbers, then
the number of prime factors of ??
??
× ??
??
×
is (x + y + z). ??
??
Recurring Decimal
Recurring decimals are referred to as
numbers that are uniformly repeated
after the decimal. Some rational
numbers produce recurring decimals
after converting them into decimal
numbers, but all irrational numbers
produce recurring decimals after
converting them into decimal form.
Examples :
(1) = 0.3333333 ….. = 0.
1
3
3
(2) 0. = = 1 9
9
9
(3) 0.53 = = 27
5327 - 53
9900
5274
9900
(4) 2.53 = 2 + = 2 27
5327 - 53
9900
5274
9900
Divisibility Test
By 2:- When last digit is 0 or an even
number eg: 520, 588
By 3:- Sum of digits is divisible by 3
eg: 1971, 1974
By 4:- When last two digits are divisible
by 4 or, they are zeros eg: 1528, 1700
By 5 :- When last digit is 0 or 5
eg: 1725, 1790
By 6 :- When the number is divisible by 2
and 3 both. eg: 36, 72
By 7 : - Subtract twice the last digit from
the number formed by the remaining
digits. Like 651 divisible by 7
65 - (1 × 2) = 63. Since 63 is divisible by
7, so is 651.
By 8 :- When the last three digits are
divisible by 8. eg: 2256
By 9 :- When sum of digit is divisible by 9
eg: 9216
By 10 :- When the last digit is 0. eg:
452600
By 11:- When the difference between the
sum of odd and even place digits is
equal to 0 or multiple of 11 .
eg: 217382
Sum of odd place digits = 2 + 7 + 8 = 17
Sum of even place digits = 1 + 3 + 2 = 6
17 – 6 = 11, hence 217382 is divisible by
11.
By 13 : - If adding four times the last
digit to the number formed by the
remaining digits is divisible by 13, then
the number is divisible by 13. Like 1326
is divisible by 13
132 + (6 4) = 156. Repeat the same ×
process for 156 .
15 + (6 4) = 39.so 39 is divisible by 13 ×
BY 17 :- The divisibility rule of 17 states,
"If ?ve times the last digit is subtracted
from a number made up of the remaining
digits and the remainder is either 0 or a
multiple of 17, then the number is
divisible by 17".
Like 221: 22 - 1 × 5 = 17.
Prime Number Test
For ?nding whether any number is a
prime number or not, we need to ?nd the
nearest square root of given number,
then we need to ?nd out whether the
given number is divisible by any prime
number less than the obtained number or
not. If it is divisible then it is not a prime
number and if not divisible then it is a
prime number.
Example : Find whether 177 is a prime
number or not.
Soln : Nearest square root of 177 is 13.
Now we need to check whether 177 is
divisible by prime numbers less than 13.
On checking we ?nd that 177 is divisible
by 3. Hence, 177 is not a prime number.
Important Formulas
1. Sum of ?rst n natural number
s =
?? ( ?? + 1 )
2
2. Sum of ?rst n odd numbers = ??
2
3. Sum of ?rst n even numbers
= ?? ( ?? + 1 )
4. Sum of square of ?rst n natural
numbers =
?? ( ?? + 1 )( 2 ?? + 1 )
6
5. Sum of cubes of ?rst n natural
number = ( )
2
?? ( ?? + 1 )
2
6. is divisible by for ( ??
??
- ??
??
) ?? - ?? ( )
every natural number m.
7. is divisible by and ( ??
??
- ??
??
) ?? + ?? ( )
for even values of m. ?? - ?? ( )
8. is divisible by for ( ??
??
+ ??
??
) ?? + ?? ( )
odd values of m.
9. Number of prime factors of
is when a, ??
??
, ??
??
, ??
??
, ??
??
?? + ?? + ?? + ??
b, c, d are all prime numbers.
10. HCF of and = ( ??
??
- 1 ) ( ??
??
- 1 )
[ ] ( ??
?????? ( ?? , ?? )
- 1 )
Number of Zeros in an expression
We shall understand this concept with
the help of an example.
Let’s ?nd the number of zeros in the
following expression: 24 32 17 23 × × ×
19 = (2
3
× 3
1
) 2
5
× 17 × 23 × 19 × ×
Notice that a zero is made only when
there is a combination of 2 and 5. Since
there is no ‘5’ here there will be no zero in
the above expression.
Example:-
8 × 15 × 23 × 17 × 25 × 22 =
2
3
× ( 3
1
× 5
1
) × 23 × 17 × 5
2
× 2
1
× 11
In this expression there are 4 twos and 3
?ves. From this 3 pairs of can be 5×2
formed. Therefore, there will be 3 zeros
in the ?nal product.
Que . Find the number of zeros in the
value of:
2
2
× 5
4
× 4
6
× 10
8
× 6
10
× 15
12
× 8
14
. × 20
16
× 10
18
× 25
20
Sol:-
2
2
× 5
4
× 4
6
× 10
8
× 6
10
× 15
12
× 8
14
= × 20
16
× 10
18
× 25
20
2
2
× 5
4
× 2
12
× 2
8
× 5
8
× 2
10
× 3
10
× 3
12
× 5
12
× 2
42
× 2
32
× 5
16
× 2
18
× 5
18
× 5
40
Zeros are possible with a combination of
2 Here the number of 5’s are less so × 5
the number of zeros will be limited to the
number of 5’s.
In this expression number of ?ves are:
5
4
× 5
8
× 5
12
× 5
16
× 5
18
× 5
40
;
i.e. 4 + 8 + 12 + 16 + 18 + 40 = 98
The number of Zeros in n!
To ?nd the number of zeros in n!, we
divide “n” by 5 until we get a number less
than 5, and then we add all the quotients
so obtained.
Que. Find the number of zeros in 36! .
The number of zeros = 7 + 1 = 8.
Remainder Theorem
Que. What will be the remainder when
is divided by 12? 17 × 23
Ans :- We can express this as:
= 17 × 23 12 + 5 ( ) × 12 + 11 ( )
= 12 × 12 + 12 × 11 + 5 × 12 + 5 × 11
In the above expression we will ?nd that
remainder will depend on the last term
i.e. 5 × 11
Now, .( ) 7. ??????
5 × 11
12
=
So,
12 × 12 + 12 × 11 + 5 × 12 + 5 × 11
12
and remainder is same in both
5 × 11
12
cases which is 7.
Example:- Remainder when
is divided by 12? 1421 × 1423 × 1425
( ) ??????
1421 × 1423 × 1425
12
= ( ) ( ) ??????
5 × 7 × 9
12
= ??????
35 × 9
12
( ) = ??????
11 × 9
12
= 3
Negative Remainder
Taking a negative remainder will make
our calculation easier.
Examples
(i) rem ( ) = rem ( )
7 × 8
9
- 2 × - 1
9
= 2 1 = 2 - × -
(ii) rem ( ) = rem ( )
55 × 56
57
- 2 × - 1
57
= - 2 1 × - = 2
(iii) rem ( ) = rem ( )
7 × 10
9
- 2 × 1
9
= 2 1 - × = - 2 ???? , 7
Large Power Concepts
Look at the following examples:
(i) rem ( ) = ( )
28
12345
9
??????
27 + 1 ( )
12345
9
= rem ( ) =
1
12345
9
1
12345
= 1
(ii) rem ( )
26
12345
9
= ( ) ??????
27 - 1 ( )
12345
9
= rem ( ) = - = - 1 or 8
- 1
12345
9
1
12345
Application of Remainder
Theorem
Que . Find the last two digits of the
expression
? 22 × 31 × 44 × 27 × 37 × 43
Sol:- If we divide the above expression by
100, we will get the last two digits as
remainder.
( ) , ? ??????
22 × 31 × 44 × 27 × 37 × 43
100
dividing by 4 to make it simple
( ) = ??????
22 × 31 × 11 × 27 × 37 × 43
25
( ) = ??????
132 × 22 × 216
25
( ) ( ) = ??????
7 × 22 × 16
25
? ??????
4 × 16
25
( ) = ??????
14
25
= 14
Since we had divided by 4 initially now to
get the correct answer, we need to
multiply the remainder by 4.
So remainder will be which 14 × 4 = 56 ,
will also be the last two digits of the
expression.
Variety Questions
Q.1. A six-digit number 11p9q4 is
divisible by 24. Then the greatest
possible value for pq is:
SSC CGL Tier II (26/10/2023)
(a) 56 (b) 68 (c) 42 (d) 32
Q.2. The remainder of the term
when divided 9 + 9
2
+............. + 9
( 2 ?? + 1 )
by 6 is:
SSC CHSL 11/08/2023 (4th Shift)
(a) 1 (b) 4 (c) 2 (d) 3
Q.3. Two numbers, when divided by a
certain divisor, leave the remainder 57.
When sum of the two numbers is divided
by the same divisor, the remainder is 49.
The divisor is:
SSC CHSL 08/08/2023 (3rd Shift)
(a) 56 (b) 57 (c) 49 (d) 65
Q.4. In a division sum, the divisor is 11
times the quotient and 5 times the
remainder. If the remainder is 44, then
the dividend is:
SSC CHSL 07/08/2023 (4th Shift)
(a) 8888 (b) 4448 (c) 8444 (d) 4444
Q.5. What is the least value of x + y, if 10
digit number 780x533y24 is divisible by
88 ?
SSC CHSL 03/08/2023 (4th Shift)
(a) 4 (b) 3 (c) 1 (d) 2
Q.6. During a division, Pranjal mistakenly
took as the dividend a number that was
10% more than the original dividend. He
also mistakenly took as the divisor a
number that was 25% more than the
original divisor. If the correct quotient of
the original division problem was 25 and
the remainder was 0, what was the
quotient that Pranjal obtained, assuming
his calculations had no error ?
SSC CGL 17/07/2023 (4th shift)
(a) 21.75 (b) 21.25 (c) 28.75 (d) 22
Q.7. The six-digit number 7x1yyx is a
multiple of 33 for non-zero digits x and y.
Which of the following could be a
possible value of (x + y) ?
Matriculation Level 30/06/2023 (Shift - 4)
(a) 5 (b) 4 (c) 2 (d) 3
Q.8. A girl wants to plant trees in her
garden in rows in such a way that the
number of trees in each row to be the
same. There are 10 rows and the number
of trees in each row is 12, what is the
number of trees in each row, if there are
5 more rows ?
SSC MTS 17/05/2023 (Evening)
(a) 10 (b) 8 (c) 6 (d) 12
Q.9. What is the total number of factors
of the number 720 except 1 and the
number itself?
SSC CHSL 10/03/2023 (4th Shift)
(a) 29 (b) 27 (c) 32 (d) 28
Q.10. Which of the following is the
smallest among ( 14 ) , ( 12 ) , ( 16 )
1
3
1
2
1
6
&
( 25 ) ?
1
12
SSC CHSL 10/03/2023 (3rd Shift)
(a) ( 14 ) (b) ( 25 ) (c) ( 16 ) (d) ( 12 )
1
3
1
12
1
6
1
2
Q.11. Which of the following statements
is correct ?
I.The Value of 100
2
- 99
2
+ 98
2
- 97
2
+ 96
2
- 95
2
+ 94
2
- 93
2
+ ...... + 22
2
- 21
2
is 4840.
II. The value of ( ) ( ) ( ??
2
+
1
??
2
?? -
1
??
??
4
)( )( ) is K
16
- +
1
??
4
?? +
1
??
??
4
-
1
??
4
1
??
16
SSC CGL 13/12/2022 (3rd Shift)
(a) Neither I nor II (b) Both I and II
(c) Only II (d) Only I
Q.12. If the seven-digit number 52A6B7C
is divisible by 33, and A, B, C are primes,
then the maximum value of 2A+3B+C is:
SSC CGL 12/12/2022 (3rd Shift)
(a) 32 (b) 23 (c) 27 (d) 34
Q.13. If the 9 - digit number 83P93678Q
is divisible by 72, then what is the value
of ? ??
2
+ ??
2
+ 12
SSC CGL 05/12/2022 (4th Shift)
(a) 6 (b) 7 (c) 8 (d) 9
Q.14. In a test (+ 5) marks are given for
every correct answer and (-2) marks are
given for every incorrect answer. Jay
answered all the questions and scored
(-12) marks, though he got 4 correct
answers. How many of his answers were
INCORRECT ?
SSC CPO 11/11/2022 (Evening)
(a) 8 (b) 32 (c) 16 (d) 20
Q.15. What is the sum of all the common
terms between the given series S1 and
S2 ?
S1 = 2, 9, 16, ……., 632
S2 = 7, 11, 15, ……., 743
SSC CGL Tier II (08/08/2022)
(a) 6974 (b) 6750 (c) 7140 (d) 6860
Q.16. If the 7 - digit number x8942y4 is
divisible by 56, what is the value of ( + ??
2
y) for the largest value of y, where x and y
are natural numbers?
SSC CGL 11/04/2022 (Evening)
(a) 33 (b) 44 (c) 55 (d) 70
Q.17. Let p, q, r and s be positive natural
numbers having three exact factors
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Pinnacle Day: 1st - 5th Number System
including 1 and the number itself If q > p
and both are two-digit numbers, and r > s
and both are one-digit numbers, then the
value of the expression is:
?? - ?? - 1
?? - ??
SSC CGL Tier II (03/02/2022)
(a) - s - 1 (b) s - 1 (c) 1 - s (d) s + 1
Q.18. Three fractions x, y and z are such
that x > y > z. When the smallest of them
is divided by the greatest, the result is
which exceeds y by 0.0625. If x + y +
9
16
z = 2 , then what is the value of x + z ?
3
12
SSC CGL Tier II (29/01/2022)
(a) (b) (c) (d)
5
4
1
4
7
4
3
4
Q.19. The six-digit number 537xy5 is
divisible by 125. How many such six -
digit numbers are there?
SSC CHSL 19/04/2021 (Morning)
(a) 4 (b) 2 (c) 3 (d) 5
Q.20. How many numbers between 400
and 700 are divisible by 5, 6 and 7 ?
SSC CPO 24/11/2020 (Evening)
(a) 2 (b) 5 (c) 10 (d) 20
Q.21. Find the number of prime factors in
the product . ( 30 )
5
× ( 24 )
5
SSC CGL Tier II (18/11/2020)
(a) 45 (b) 35 (c) 10 (d) 30
Q.22. Let ab, a b, is a 2-digit prime ?
number such that ba is also a prime
number. The sum of all such number is:
SSC CGL Tier II (16/11/2020)
(a) 374 (b) 418 (c) 407 (d) 396
Q.23. Given that 2
20
+ 1 is completely
divisible by a whole number. Which of the
following is completely divisible by the
same number ?
SSC CHSL 16/10/2020 (Afternoon)
(a) 2
15
+ 1 (b) 5 2
30
×
(c) 2
90
+ 1 (d) 2
60
+ 1
Q.24. Which of the following numbers
will completely divide 7
81
+ 7
82
+ 7
83
?
SSC CHSL 17/03/2020 (Morning)
(a) 399 (b) 389 (c) 387 (d) 397
Q.25. When a positive integer is divided
by d, the remainder is 15. When ten times
of the same number is divided by d, the
remainder is 6. The least possible value
of d is:
SSC CGL 05/03/2020 (Afternoon)
(a) 9 (b) 12 (c) 16 (d) 18
Q.26. When 200 is divided by a positive
integer x, the remainder is 8. How many
values of x are there?
SSC CGL 03/03/2020 (Afternoon)
(a) 7 (b) 5 (c) 8 (d) 6
Q.27. The number 1563241234351 is :
SSC CPO 13/12/2019 (Evening)
(a) divisible by both 3 and 11
(b) divisible by 11 but not by 3
(c) neither divisible by 3 nor by 11
(d) divisible by 3 but not by 11
Q.28. How many natural numbers less
than 1000 are divisible by 5 or 7 but NOT
by 35 ?
SSC CPO 11/12/2019 (Morning)
(a) 285 (b) 313 (c) 341 (d) 243
Q.29. If r is the remainder when each of
4749, 5601 and 7092 is divided by the
greatest possible number d ( 1), then >
the value of (d + r) will be:
SSC CPO 11/12/2019 (Morning)
(a) 276 (b) 271 (c) 298 (d) 282
Q.30. Let x be the least 4-digit number
which when divided by 2, 3, 4, 5, 6 and 7
leaves a remainder of 1 in each case. If x
lies between 2800 and 3000, then what is
the sum of digits of x ?
SSC CPO 09/12/2019 (Evening)
(a) 15 (b) 16 (c) 12 (d) 13
Q.31. If the six digit number 479xyz is
exactly divisible by 7, 11 and 13 , then {( y
+ z ) ÷ x} is equal to :
SSC CPO 09/12/2019 (Morning)
(a) (b) 4 (c) (d)
11
9
13
7
7
13
Q.32. Which among the following is the
smallest?
SSC CPO 09/12/19 (Morning)
(a) v401 - v399 (b) v101 - v99
(c) v301 - v299 (d) v201 - v199
Q.33. If x is the remainder when 3
61284
is
divided by 5 and y is the remainder when
4
96
is divided by 6, then what is the value
of (2x - y) ?
SSC CGL Tier II (13/09/2019)
(a) - 4 (b) 4 (c) -2 (d) 2
Q.34. In ?nding the HCF of two numbers
by division method, the last divisor is 17
and the quotients are 1, 11 and 2,
respectively. What is the sum of the two
numbers ?
SSC CGL Tier II (13/09/2019)
(a) 833 (b) 867 (c) 816 (d) 901
Q.35. Two positive numbers differ by
2001. When the larger number is divided
by the smaller number, the quotient is 9
and the remainder is 41. The sum of the
digits of the larger number is :
SSC CGL Tier II (13/09/2019)
(a) 15 (b) 11 (c) 10 (d) 14
Q.36. When a two-digit number is
multiplied by the sum of its digits, the
product is 424. When the number
obtained by interchanging its digits is
multiplied by the sum of the digits, the
result is 280. The sum of the digits of the
given number is :
SSC CGL Tier II (12/09/2019)
(a) 6 (b) 9 (c) 8 (d) 7
Q.37. Let x be the least number which
when divided by 15,18,20 and 27, the
remainder in each case is 10 and x is a
multiple of 31. What least number should
be added to x to make it a perfect square ?
SSC CGL Tier II (12/09/2019)
(a) 39 (b) 37 (c) 43 (d) 36
Q.38. The number of factors of 3600 is :
SSC CGL Tier II (12/09/2019)
(a) 45 (b) 44 (c) 43 (d) 42
Q.39. When 12, 16, 18, 20 and 25 divide
the least number x, the remainder in each
case is 4 but x is divisible by 7. What is
the digit at the thousands’ place in x ?
SSC CGL Tier II (11/09/2019)
(a) 5 (b) 8 (c) 4 (d) 3
Q.40. One of the factors of ( ), 8
2 ??
+ 5
2 ??
where k is an odd number, is :
SSC CGL Tier II (11/09/2019)
(a) 86 (b) 88 (c) 84 (d) 89
Q.41. Let x = ( 633 )
24
- ( 277 )
38
+
What is the unit digit of x ? ( 266 )
54
SSC CGL Tier II (11/09/2019)
(a) 7 (b) 6 (c) 4 (d) 8
Q.42. The sum of the digits of a two-digit
number is of the number. The unit
1
7
digit is 4 less than the tens digit. If the
number obtained on reversing its digit is
divided by 7, the remainder will be :
SSC CGL Tier II (11/09/2019)
(a) 4 (b) 5 (c) 1 (d) 6
Q.43. When 6892, 7105 and 7531 are
divided by the greatest number x, then
the remainder in each case is y. What is
the value of (x - y) ?
SSC MTS 22/08/2019 (Afternoon)
(a) 123 (b) 137 (c) 147 (d) 113
Q.44. Let x be the greatest number which
when divides 6475, 4984 and 4132, the
remainder in each case is the same.
What is the sum of digits of x?
SSC MTS 22/08/2019 (Morning)
(a) 4 (b) 7 (c) 5 (d) 6
Q.45. When an integer n is divided by 8,
the remainder is 3. What will be the
remainder if 6n - 1 is divided by 8 ?
SSC CGL 13/06/2019 (Evening)
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Pinnacle Day: 1st - 5th Number System
(a) 4 (b) 1 (c) 0 (d) 2
Q.46. If a 11 digit number 5y5884805x6
is divisible by 72, where x = y, then the
value of is : ????
SSC CGL 10/06/2019 (Morning)
(a) (b) 3 (c) 7 (d) 2 7 7
Q.47. A gardener planted 1936 saplings
in a garden such that there were as many
rows of saplings as the columns. The
number of rows planted is :
SSC CPO 16/03/2019 (Afternoon)
(a) 46 (b) 44 (c) 48 (d)42
Q.48. What is the difference between the
largest and smallest numbers of the four
digits created using numbers 2, 9, 6, 5 ?
(Each number can be used only once)
SSC CPO 14/03/2019 (Evening)
(a) 6993 (b) 7056 (c) 6606 (d) 7083
Q.49. The sum of all possible three digit
numbers formed by digits 3, 0 and 7,
using each digit only once is:
SSC CPO 14/03/2019 (Morning)
(a) 2010 (b) 1990 (c) 2220 (d) 2110
Q.50. What is the sum of digits of the
least number, which when divided by 15,
18 and 42 leaves the same remainder 8
in each case and is also divisible by 13 ?
SSC CPO 13/03/2019 (Evening)
(a) 25 (b) 24 (c) 22 (d) 26
Q.51. The square root of which of the
following is a rational number ?
SSC CPO 12/03/2019 (Morning)
(a) 1250.49 (b) 6250.49
(c) 1354.24 (d) 5768.28
Q.52. What is the sum of the digits of the
least number, which when divided by 12 ,
16 and 54, leaves the same remainder 7
in each case and is also completely
divisible by 13 ?
SSC CPO 12/03/2019 (Evening)
(a) 36 (b) 16 (c) 9 (d) 27
Practice Questions
SSC CHSL 2023 Tier - 1
Q.53. Which of the following is the
nearest number to 13051 and is divisible
by 9 ?
SSC CHSL 02/08/2023 (3rd Shift)
(a) 13057 (b) 13056 (c) 13059 (d) 13058
Q.54. A Number when divided by 78
gives the quotient 280 and the remainder
0. If the same number is divided by 65,
what will be the value of the remainder ?
SSC CHSL 02/08/2023 (4th Shift)
(a) 1 (b) 3 (c) 0 (d) 2
Q.55. The divisor is 10 times the quotient
and 5 times the remainder in a division
sum. What is the dividend if the
remainder is 46 ?
SSC CHSL 03/08/2023 (2nd Shift)
(a) 5972 (b) 4286 (c) 4874 (d) 5336
Q.56. What is the largest ?ve digit
number exactly divisible by 88 ?
SSC CHSL 04/08/2023 (1st Shift)
(a) 99992 (b) 99986 (c) 99984 (d) 99968
Q.57. What is the least number that must
be added to the greatest 6-digit number
so that the sum will be exactly divisible
by 294 ?
SSC CHSL 07/08/2023 (2nd Shift)
(a) 234 (b) 194 (c) 269 (d) 189
Q.58. Which of the following sets is such
that all its elements are divisors of the
number 2520 ?
SSC CHSL 08/08/2023 (1st Shift)
(a) 12, 49, 18 (b) 8 , 9, 7
(c) 16, 15, 14 (d) 21, 10, 25
Q.59. What is the remainder when 3
8
is
divided by 7 ?
SSC CHSL 08/08/2023 (2nd Shift)
(a) 5 (b) 4 (c) 6 (d) 2
Q.60. If 7 divides the integer n, then the
remainder is 2. What will be the
remainder if 9n is divided by 7 ?
SSC CHSL 09/08/2023 (1st Shift)
(a) 3 (b) 5 (c) 1 (d) 4
Q.61. What will be the greatest number
32a78b, which is divisible by 3 but NOT
divisible by 9 ? (Where a and b are single
digit numbers).
SSC CHSL 09/08/2023 (2nd Shift)
(a) 324781 (b) 329787
(c) 326787 (d) 329784
Q.62. The sum of the cubes of two given
numbers is 10234, while the sum of the
two given numbers is 34. What is the
positive difference between the cubes of
the two given numbers ?
SSC CHSL 11/08/2023 (1st Shift)
(a) 3484 (b) 3488 (c) 3356 (d) 8602
Q.63. If a 10-digit number 620x976y52 is
divisible by 88, then the least value of (x²
+ y²) will be:
SSC CHSL 14/08/2023 (3rd Shift)
(a) 8 (b) 7 (c) 11 (d) 10
Q.64. The six-digit number N = 4a6b9c is
divisible by 99, then the maximum sum
of the digits of N is:
SSC CHSL 17/08/2023 (1st Shift)
(a) 18 (b) 36 (c) 45 (d) 27
SSC CGL 2023 Tier - 1
Q.65. What will be the remainder when
(265)
4081
+ 9 is divided by 266 ?
SSC CGL 14/07/2023 (1st shift)
(a) 8 (b) 6 (c) 1 (d) 9
Q.66. A and B have some toffees. If A
gives one toffee to B, then they have an
equal number of toffees. If B gives one
toffee to A, then the toffees with A are
double with B. The total number of
toffees with A and B are _______.
SSC CGL 14/07/2023 (3rd shift)
(a) 12 (b) 10 (c) 14 (d) 15
Q.67. Find the smallest number that can
be subtracted from 148109326 so that it
becomes divisible by 8.
SSC CGL 17/07/2023 (1st shift)
(a) 4 (b) 8 (c) 6 (d) 10
Q.68. The largest 5-digit number exactly
divisible by 88 is:
SSC CGL 17/07/2023 (2nd shift)
(a) 99990 (b) 99984 (c) 99978 (d) 99968
Q.69. Find the sum of 3 + 3
2
+ 3
3
+ ……. + 3
8
.
SSC CGL 17/07/2023 (2nd shift)
(a) 6561 (b) 6560 (c) 9840 (d) 3280
Q.70. How many of the following
numbers are divisible by 132 ?
660, 754, 924, 1452, 1526, 1980, 2045
and 2170
SSC CGL 17/07/2023 (3rd shift)
(a) 3 (b) 6 (c) 5 (d) 4
Q.71. A six - digit number is divisible by
33. If 54 is added to the number, then the
new number formed will also be divisible
by :
SSC CGL 17/07/2023 (4th shift)
(a) 3 (b) 2 (c) 5 (d) 7
Q.72. Which of the following numbers is
divisible by 24 ?
SSC CGL 18/07/2023 (1st shift)
(a) 52668 (b) 49512 (c) 64760 (d) 26968
Q.73. The cost of 32 pens and 12 pencils
is ?790. What is the total cost (in ? ) of 8
pens and 3 pencils together?
SSC CGL 18/07/2023 (2nd shift)
(a) 200.5 (b) 197.5 (c) 180.5 (d) 220.5
Q.74. The smallest number added to 888
so that it is exactly divisible by 35 is :
SSC CGL 18/07/2023 (3rd shift)
(a) 22 (b) 23 (c) 20 (d) 21
Q.75. An 11-digit number 7823326867X
is divisible by 18. What is the value of X?
SSC CGL 19/07/2023 (1st shift)
(a) 6 (b) 4 (c) 8 (d) 2
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FAQs on SSC CGL Previous Year Questions (2023-18): Number System - SSC CGL Mathematics Previous Year Paper (Topic-wise)
| 1. What are the different types of numbers in the number system? |  |
Ans. The number system includes various types of numbers such as natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers.
| 2. How can one determine if a number is prime or composite? | |
Ans. A number is considered prime if it has exactly two factors, 1 and itself. If a number has more than two factors, then it is classified as a composite number.
| 3. What is the significance of the concept of LCM and HCF in the number system? | |
Ans. The concept of Least Common Multiple (LCM) and Highest Common Factor (HCF) is crucial in simplifying fractions, solving word problems, and finding the common multiple or factor of two or more numbers.
| 4. How can one convert a decimal number to a fraction in the number system? | |
Ans. To convert a decimal number to a fraction, count the number of decimal places and put the decimal number over a power of 10 (e.g., 0.5 = 5/10 = 1/2).
| 5. How do you find the square root of a number in the number system? | |
Ans. To find the square root of a number, one can use the prime factorization method, long division method, or estimation method depending on the complexity of the number.
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Additional Information about SSC CGL Previous Year Questions (2023-18): Number System for SSC CGL Preparation
SSC CGL Previous Year Questions (2023-18): Number System Free PDF Download
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SSC CGL Previous Year Questions (2023-18): Number System Notes
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SSC CGL Previous Year Questions (2023-18): Number System SSC CGL
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The content of SSC CGL Previous Year Questions (2023-18): Number System is prepared as per the latest SSC CGL syllabus.
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