SSC CGL Previous Year Question: Number System and HCF & LCM (2023-2024)

If a number is divisible by 3 then its sum should also be divisible by 3.
5 + 3 + 0 + 6 + P + 2 = 16 + P
Least value = P = 2 as 16 + 2 = 18 (divisible by 3)
Maximum value P = 8 as 16 + 8 = 24 (divisible by 3)
Difference in Square = 64 - 4 = 60

72 = 8 × 9
For divisibility of 8, last 3 digits should be divisible by 8
Possible value of y = 3 and 7
For divisibility of 9, the sum of digits should be divisible by 9.
9 + 4 + x + 2 + 9 + 3 + 6 = 9 or its multiple x = 3 (x = y)
Check with 7
9 + 4 + x + 2 + 9 + 7 + 6 = 9
or its multiple x = 8
So , 2x + 3y = 2(8) + 3(7) = 37

For divisibility of 3, the sum of digits should be divisible by 3
4 + 5 + 0 + 8 + 2 + k = multiple of 3
19 + k = multiple of 3
Possible value of k = 2, 5, 8
Required Answer = 8² + 2² = 68

P = 2x , Q = 3y
P × Q = 6xy
6xy is divisible by 6
P × Q is also divisible by 6 but not 5
P + Q = 2x + 3y
2x + 3y is not divisible by 5 and 6
So, P + Q is not divisible by 5 and 6.

Let five consecutive odd natural numbers are
(x - 4), (x - 2), x, (x + 2), (x + 4)
(x - 4)² + (x - 2)² + x² + (x + 2)² + (x + 4)² = 233 x 5 ⇒ 5x² + 40 = 1165
5x² = 1125 ⇒ x = 15,
Largest number = 19
Smallest number = 11
Average = 15

LCM of 3, 7, 11 = 231
When 59399 is divided by 231, Reminder = 32
Required number = 59399 - 32 = 59367
Compare 59367 with 593ab
so, a = 6 , b = 7 , (a² - b² + ab) = 29

If a number is divisible by 7
11 and 13 then it must be divisible by 1001
And if the number is divisible by 1001 the its first three digits would be same as its next three digits (xyzxyz)
So, z = 4, x = 5 and y = 3 and x + y - z = 8 - 4 = 4.

4a067b
Sum of odd placed digits = 4 + 0 + 7
Sum of even placed digits = a + 6 + b
So, 11 - (6 + a + b) = 0 or multiple of 11
a + b = 5 or 16
So the sum all the possible values of (a + b) = 5 + 16 = 21

(4x - 2y + 3z)
= 4 × 17 − 24 × 2 + 27 × 3
(4x - 2y + 3z) = 101
When 101 is divided by 31 we get reminder = 8

LCM of 3, 7, 11 is = 231
Let the number be 23599
When 23599 is divided by 231 we get remainder as 37
So the number is = 23599 - 37 = 23562
x = 6 and y = 2
(3x - 4y) = 18 - 8 = 10

As 10404 is a perfect square
So x = 20
Prime factorization of 20 = 2 × 2 × 5
So, when 20 is multiplied by 5 it become a perfect square

a = 9, b = 7 and c = 10
(a + 2b + 5c) = (9 + 14 + 50) = 73
When 73 is divided by 13 remainder is 8.

When 732 is divided by x, then the remainder is 12. So, the number exactly divisible by x is 720.
720 = 2⁴ x 3² x 5¹
Now, the no of factors of 720 = (4 + 1) × (2 + 1) × (1 + 1) = 30
The remainder is always less than the divisor , so x > 12
So, the no of factors of 720 which are less than 12 = (1 , 2 , 3 , 4 , 5 , 6 , 8 , 9 , 10, 12) = 10
So, possible values x = 30 - 10 = 20.

n = 7Q + 2
For remainder 0, add 5 both sides, we get: n + 5 = 7Q + 7

Let the number be x which divides 7897, 8110 and 8536 leaving a reminder r.
The required number then becomes H.C.F of (7897 - r), (8110 - r) and (8536 - r)
It could also be the H.C.F of (8536 - r) - (8110 - r) and (8110 - r) - (7897 - r) .i.e. 426 and 213
⇒ H.C.F of 426 and 213 = 213 the required sum = 2 + 1 + 3 = 6

Given, a + b + c = 1(11/12) ...(1) and (c/a) = 5/2
Let c = 5 unit and a = 2 unit
According to the question
b = (5/2) - (7/4) = 3/4
Put this value in eq (1)
a + c = (23/12) - (3/4) = 7/6
(5 + 2) unit = 7/6 ⇒ 1 unit = 1/6
⇒ 5 unit = 5/6 ⇒ 2 unit = 1/3
Required difference = 5/6 - 1/3 = 1/2.

Given,
x + y + z = 1(13/24) ...(1) and z/x = 9/16
Let z = 9 unit and x = 16 unit
According to the question
y = (9/16) - 0.0625
= (9/16) - (625/10000) = 1/2
Put this value in eq (1)
⇒ x + z = 

Given,
x = (164)¹⁶⁹ + (333)³³⁷ - (727)⁷²⁶
Unit digit of (164)¹⁶⁹ = 4¹ = 4
Unit digit of (333)³³⁷ = 3¹ = 3
Unit digit of (727)⁷²⁶ = 7² = 9
Unit digit of x = 4 + 3 - 9 = - 2
But unit digit can’t be negative so, required unit digit = 10 + (-2) = 8

Since the number is divisible by 72 it must be divisible by 9 and 8.
The number to be divisible by 8, the last three digits must be divisible by 8.
When 7x² is divided by 8, the possible value for x are = 1, 5, 9
The number to If x = 1, then y = 3
If x = 5, then y = 8 be divisible by 9, the sum of numbers must be divisible by 9.
8 + 9 + 7 + 3 + 5 + 9 + y + 7 + x + 2 = divisible by 9
50 + x + y = 9
If x = 9, then y = 4
If y = 8 (greatest value), then x = 5
3x - y = 3 (5) - 8 = 7

Number 15x1y2 is divisible by 44, clearly it will also be divisible by 11 and 4 .
A number to be divisible by 11
(1 + x + y ) - ( 5 + 1 + 2) = 0 or 11
For the difference = 0 ⇒ x + y = 7
For the difference = 11 ⇒ x + y = 18
But 18 is not given in the options so option (b) is the right answer.

Since, 517X324 is divisible by 12 it must be divisible by 3 and 4(coprime factors of 12).
For a number to be divisible by 3, the sum of its digits must be divisible by 3.
So, 5 + 1 + 7 + x + 3 + 2 + 4 ⇒ 22 + x must be divisible by 3
Possible values of x = 2, 5, 8
Cleary the smallest value of x = 2

L.C.M of 2, 3, 5, 6, 9 and 18 = 90
90 = 2 × 3 × 3 × 5
As 2 and 5 don’t have any pair
So, the required number = 90 × 2 × 5 = 900

Let, the no. are 15a and 15b.
36x² - 81y² = 9(2x -3y)(2x + 3y)
= 9(2 x 15a - 3 x 15b)(2 x 15a + 3 x 15b)
= 9 x 15{(2a - 3b)(2a + 3b)}
And, 81x² - 9y² = 9(3x - y)(3x + y)
= 9 (3 x 15a − 15b)(3 x 15a + 15b)
= 9 x 15 {(3a − b)(3a + b)}
Their H.C.F.= 135

H.C.F. 

H.C.F of two or more numbers is the highest number which exactly divides all the given numbers.
Example:- H.C.F of 4 and 6 = 2
2 exactly divides 4 and 6.

LCM (8, 9, 12, 15) = 360
So, the number is 360k + 5, which is divisible by 7 ⇒ Putting k = 3, we have ;
Required no = 360 × 3 + 5 = 1080 + 5 = 1085

Let the LCM and HCF of the two no’s be L and H respectively
Then, L + H = 512 ---------(1)
and, L - H = 496 -----------(2)
Adding eqn (1) and (2) we have :

Now, putting the value of L in equation (1),
H = 512 - 504 = 8
Product of two numbers = LCM × HCF
⇒ 72 × y = 8 × 504

According to the question,

Let three numbers be 6x, 8x and 9x
So, 9x – 6x = 33 ⇒ 3x = 33 ⇒ x = 11
HCF (6x, 8x, 9x) = x = 11

To find the required greatest number, we need to find the HCF of the difference between all the three given
numbers.
1134 - 1062 = 72, 1182 - 1134 = 48
HCF of 72 and 48 = 24
Now, When 1062 is divided by 24 leaves, remainder 6.
So, x - y = 24 - 6 = 18

To find the value of x, we need to find the HCF of the difference between all the three given numbers.
1027 - 955 = 72, 1075 - 1027 = 48
HCF of (72 and 48) = 24
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Hence, 16 is not the factor of x.