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All questions of Remainder & Divisibility for Interview Preparation Exam
After the division of a number successively by 3, 4 and 7, the remainders obtained are 2,1 and 4 respectively. What will be the remainder if 84 divides the same number?- a)80
- b)75
- c)41
- d)53
Correct answer is option 'D'. Can you explain this answer?
After the division of a number successively by 3, 4 and 7, the remainders obtained are 2,1 and 4 respectively. What will be the remainder if 84 divides the same number?
a)
80
b)
75
c)
41
d)
53
|
Arun Sharma answered |
In the successive division, the quotient of first division becomes the dividend of the second division and so on.
Let the last quotient be p, so the last dividend will be 7p + 4 which is the quotient of the second division.
So, the second dividend is (7p + 4) � 4 + 1.
Applying the same logic, the number = 3 {4(7p + 4) + 1} + 2 = 84p + 53
Hence, if the number is divided by 84, the remainder is 53.
Let the last quotient be p, so the last dividend will be 7p + 4 which is the quotient of the second division.
So, the second dividend is (7p + 4) � 4 + 1.
Applying the same logic, the number = 3 {4(7p + 4) + 1} + 2 = 84p + 53
Hence, if the number is divided by 84, the remainder is 53.
Hence Option D is correct
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What will be the unit digit of 1341 ?- a)3
- b)0
- c)1
- d)9
Correct answer is option 'A'. Can you explain this answer?
What will be the unit digit of 1341 ?
a)
3
b)
0
c)
1
d)
9
|
|
Arun Sharma answered |
As we know the last digit depends upon the unit digit of the multiplier numbers so the unit digit of 1341
is same as the last digit of 341 and we know that the cyclicity of 3 is 4
On dividing the number 41 with 4 and we will get the remainder as 1 and the last digit will be 31 = 3
is same as the last digit of 341 and we know that the cyclicity of 3 is 4
On dividing the number 41 with 4 and we will get the remainder as 1 and the last digit will be 31 = 3
Find the maximum value of n such that 50! is perfectly divisible by 2520". - a)6
- b)8
- c)7
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Find the maximum value of n such that 50! is perfectly divisible by 2520".
a)
6
b)
8
c)
7
d)
None of these
|
Shalini Patel answered |
2520=7 X 32 X 23 X 5.
The value of n would be given by the value of the number of 7s in 50! This Value Is Equal To[50/7] + [50/49] = 7 + 1 = 8 Option (b) is correct.
What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?- a)666
- b)676
- c)683
- d)777
Correct answer is option 'B'. Can you explain this answer?
What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?
a)
666
b)
676
c)
683
d)
777
|
|
Niti Joshi answered |
To find the sum of all two-digit numbers that give a remainder of 3 when divided by 7, we need to identify all the two-digit numbers that satisfy this condition and then add them up.
Finding the two-digit numbers:
To find the two-digit numbers that give a remainder of 3 when divided by 7, we can start by listing the numbers from 10 to 99 and checking if each number satisfies the condition. However, this method can be time-consuming.
A more efficient approach is to observe the pattern of remainders when dividing numbers by 7. The remainders repeat after every 7 numbers. The remainders for the numbers 10 to 16 are 3, 4, 5, 6, 0, 1, 2. From this pattern, we can see that the numbers that give a remainder of 3 when divided by 7 are 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, 80, 87, and 94.
Summing the two-digit numbers:
To find the sum of these numbers, we can employ the formula for the sum of an arithmetic series. The formula is Sn = (n/2)(a + l), where Sn is the sum, n is the number of terms, a is the first term, and l is the last term. In this case, the first term is 10, the last term is 94, and the number of terms is 13.
Using the formula, we can calculate the sum as follows:
Sn = (13/2)(10 + 94)
= (13/2)(104)
= 13 * 52
= 676
Therefore, the sum of all two-digit numbers that give a remainder of 3 when divided by 7 is 676.
In conclusion, the correct answer is option 'B' (676).
Finding the two-digit numbers:
To find the two-digit numbers that give a remainder of 3 when divided by 7, we can start by listing the numbers from 10 to 99 and checking if each number satisfies the condition. However, this method can be time-consuming.
A more efficient approach is to observe the pattern of remainders when dividing numbers by 7. The remainders repeat after every 7 numbers. The remainders for the numbers 10 to 16 are 3, 4, 5, 6, 0, 1, 2. From this pattern, we can see that the numbers that give a remainder of 3 when divided by 7 are 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, 80, 87, and 94.
Summing the two-digit numbers:
To find the sum of these numbers, we can employ the formula for the sum of an arithmetic series. The formula is Sn = (n/2)(a + l), where Sn is the sum, n is the number of terms, a is the first term, and l is the last term. In this case, the first term is 10, the last term is 94, and the number of terms is 13.
Using the formula, we can calculate the sum as follows:
Sn = (13/2)(10 + 94)
= (13/2)(104)
= 13 * 52
= 676
Therefore, the sum of all two-digit numbers that give a remainder of 3 when divided by 7 is 676.
In conclusion, the correct answer is option 'B' (676).
Find the remainder when 73 + 75 + 78 + 57 + 197 is divided by 34.- a)32
- b)4
- c)15
- d)28
Correct answer is option 'B'. Can you explain this answer?
Find the remainder when 73 + 75 + 78 + 57 + 197 is divided by 34.
a)
32
b)
4
c)
15
d)
28
|
|
Shalini Patel answered |
The remainder would be given by: (5 + 7 + 10 + 23 + 27)/34 = 72/34
-> remainder = 4. Option (b) is correct.
-> remainder = 4. Option (b) is correct.
Let n! = 1 x 2 x 3 x……….x n for integer n ≥ 1. If p = 1! + (2 x 2!) + (3 x 3!) + ……(10 x 10!), then p + 2 when divided by 11! Leaves remainder of- a)10
- b)0
- c)7
- d)1
Correct answer is option 'D'. Can you explain this answer?
Let n! = 1 x 2 x 3 x……….x n for integer n ≥ 1. If p = 1! + (2 x 2!) + (3 x 3!) + ……(10 x 10!), then p + 2 when divided by 11! Leaves remainder of
a)
10
b)
0
c)
7
d)
1
|
|
Arun Sharma answered |
If P = 1! = 1
Then P + 2 = 3, when divided by 2! remainder will be 1.
If P = 1! + 2 × 2! = 5
Then, P + 2 = 7 when divided by 3! remainder is still 1.
Hence, P = 1! + (2 × 2!) + (3 × 3!)+ ……+ (10 × 10!)
Hence, when p + 2 is divided by 11!, the remainder is 1.
Alternative method:
P = 1 + 2 × 2! + 3 × 3! + …..10 × 10!
= (2 – 1)1! + (3 – 1)2! + (4 – 1)3! + …+ (11 – 1)10!
= 2! – 1! + 3! – 2! + …..+ 11! – 10!
= 11! – 1
Hence p + 2 = 11! + 1
Hence, when p + 2 is divided by 11!, the remainder is 1
Then P + 2 = 3, when divided by 2! remainder will be 1.
If P = 1! + 2 × 2! = 5
Then, P + 2 = 7 when divided by 3! remainder is still 1.
Hence, P = 1! + (2 × 2!) + (3 × 3!)+ ……+ (10 × 10!)
Hence, when p + 2 is divided by 11!, the remainder is 1.
Alternative method:
P = 1 + 2 × 2! + 3 × 3! + …..10 × 10!
= (2 – 1)1! + (3 – 1)2! + (4 – 1)3! + …+ (11 – 1)10!
= 2! – 1! + 3! – 2! + …..+ 11! – 10!
= 11! – 1
Hence p + 2 = 11! + 1
Hence, when p + 2 is divided by 11!, the remainder is 1
Find the remainder when 51203 is divided by 7.- a)4
- b)2
- c)1
- d)6
Correct answer is option 'A'. Can you explain this answer?
Find the remainder when 51203 is divided by 7.
a)
4
b)
2
c)
1
d)
6
|
|
Shalini Patel answered |
51203/7 -> 2203/7 = (23)67 X 22/7 = 867 X 4/7 -> remainder = 4. Option (a) is correct
What will be the last digit of 34 x 45 x 56- a)8
- b)0
- c)9
- d)3
Correct answer is option 'B'. Can you explain this answer?
What will be the last digit of 34 x 45 x 56
a)
8
b)
0
c)
9
d)
3
|
Krishna Iyer answered |
Last digit of 34 = 1
Last digit of 45 = 4
Last digit of 56 = 5
So as we discussed earlier that the last digit depends upon the last digits
So 1 x 4 x 5 = 20 so the last digit is 0; also whenever 5 is multiplied by even
number then last digit will be 0
Last digit of 45 = 4
Last digit of 56 = 5
So as we discussed earlier that the last digit depends upon the last digits
So 1 x 4 x 5 = 20 so the last digit is 0; also whenever 5 is multiplied by even
number then last digit will be 0
Find the number of consecutive zeroes at the end of the following numbers. 72!- a)13
- b)14
- c)15
- d)16
Correct answer is option 'D'. Can you explain this answer?
Find the number of consecutive zeroes at the end of the following numbers. 72!
a)
13
b)
14
c)
15
d)
16
|
|
Shalini Patel answered |
The number of zeroes would depend on the number
of 5’s in the value of the factorial. 72! -> 14 + 2 = 16. Option (d) is correct.
Find the number of consecutive zeroes at the end of the following numbers. 100! + 200! - a)73
- b)24
- c)11
- d)22
Correct answer is option 'B'. Can you explain this answer?
Find the number of consecutive zeroes at the end of the following numbers. 100! + 200!
a)
73
b)
24
c)
11
d)
22
|
|
Pooja Nambiar answered |
Consecutive zeroes at the end of a number:
When we talk about consecutive zeroes at the end of a number, we are essentially looking for the number of factors of 10 in that number. This is because a factor of 10 represents a trailing zero at the end of a number.
Factors of 10:
To understand the factors of 10, let's break down the number 10 into its prime factors:
10 = 2 * 5
The prime factors of 10 are 2 and 5. In order to have a factor of 10, we need both 2 and 5 to be present in the prime factorization of a number.
Factors of 2 and 5:
Let's consider the prime factors of numbers from 1 to 10:
1 = 1
2 = 2
3 = 3
4 = 2 * 2
5 = 5
6 = 2 * 3
7 = 7
8 = 2 * 2 * 2
9 = 3 * 3
10 = 2 * 5
From the above list, we can observe that the number of factors of 2 is more than the number of factors of 5. Therefore, the number of factors of 10 will be limited by the number of factors of 5.
Calculating the factors of 5 in a number:
To find the number of factors of 5 in a number, we divide the number by 5 and count the number of whole numbers obtained. For example, in the range from 1 to 10, the number of factors of 5 is 1 (from 5 itself).
Calculating the factors of 10 in a number:
To find the number of factors of 10 in a number, we divide the number by 10 and count the number of whole numbers obtained. For example, in the range from 1 to 10, the number of factors of 10 is 0 (as there are no whole numbers obtained).
Calculating the factors of 5 and 10 in 100!:
In 100!, the number of factors of 5 will be determined by dividing 100 by 5 and counting the number of whole numbers obtained. Similarly, the number of factors of 10 will be determined by dividing 100 by 10 and counting the number of whole numbers obtained.
To calculate the number of factors of 5 and 10 in 100!, we can follow these steps:
1. Divide 100 by 5: 100 ÷ 5 = 20
2. Divide 100 by 10: 100 ÷ 10 = 10
Therefore, there are 20 factors of 5 and 10 factors of 10 in 100!.
Calculating the factors of 5 and 10 in 200!:
Similarly, in 200!, the number of factors of 5 will be determined by dividing 200 by 5 and counting the number of whole numbers obtained. The number of factors of 10 will be determined by dividing 200 by 10 and counting the number of whole numbers obtained.
To calculate the number of factors of 5 and 10 in 200!, we can follow these steps:
1. Divide 200 by 5
When we talk about consecutive zeroes at the end of a number, we are essentially looking for the number of factors of 10 in that number. This is because a factor of 10 represents a trailing zero at the end of a number.
Factors of 10:
To understand the factors of 10, let's break down the number 10 into its prime factors:
10 = 2 * 5
The prime factors of 10 are 2 and 5. In order to have a factor of 10, we need both 2 and 5 to be present in the prime factorization of a number.
Factors of 2 and 5:
Let's consider the prime factors of numbers from 1 to 10:
1 = 1
2 = 2
3 = 3
4 = 2 * 2
5 = 5
6 = 2 * 3
7 = 7
8 = 2 * 2 * 2
9 = 3 * 3
10 = 2 * 5
From the above list, we can observe that the number of factors of 2 is more than the number of factors of 5. Therefore, the number of factors of 10 will be limited by the number of factors of 5.
Calculating the factors of 5 in a number:
To find the number of factors of 5 in a number, we divide the number by 5 and count the number of whole numbers obtained. For example, in the range from 1 to 10, the number of factors of 5 is 1 (from 5 itself).
Calculating the factors of 10 in a number:
To find the number of factors of 10 in a number, we divide the number by 10 and count the number of whole numbers obtained. For example, in the range from 1 to 10, the number of factors of 10 is 0 (as there are no whole numbers obtained).
Calculating the factors of 5 and 10 in 100!:
In 100!, the number of factors of 5 will be determined by dividing 100 by 5 and counting the number of whole numbers obtained. Similarly, the number of factors of 10 will be determined by dividing 100 by 10 and counting the number of whole numbers obtained.
To calculate the number of factors of 5 and 10 in 100!, we can follow these steps:
1. Divide 100 by 5: 100 ÷ 5 = 20
2. Divide 100 by 10: 100 ÷ 10 = 10
Therefore, there are 20 factors of 5 and 10 factors of 10 in 100!.
Calculating the factors of 5 and 10 in 200!:
Similarly, in 200!, the number of factors of 5 will be determined by dividing 200 by 5 and counting the number of whole numbers obtained. The number of factors of 10 will be determined by dividing 200 by 10 and counting the number of whole numbers obtained.
To calculate the number of factors of 5 and 10 in 200!, we can follow these steps:
1. Divide 200 by 5
Find the number of consecutive zeroes at the end of the following numbers. - 100! x 200! - a)49
- b)73
- c)132
- d)33
Correct answer is option 'B'. Can you explain this answer?
Find the number of consecutive zeroes at the end of the following numbers. - 100! x 200!
a)
49
b)
73
c)
132
d)
33
|
|
Shalini Patel answered |
The number of zeroes would depend on the number of 5’s in the value of the factorial.
100! would end in 20 + 4 = 24 zeroes
200! Would end in 40 + 8 + 1 = 49 zeroes.
When you multiply the two numbers (one with 24 zeroes and the other with 49 zeroes at it’s end), the resultant total would end in 24 + 49 = 73 zeroes. Option (b) is correct.
Identify the last digit of (794 + 875)- a)8
- b)0
- c)7
- d)4
Correct answer is option 'A'. Can you explain this answer?
Identify the last digit of (794 + 875)
a)
8
b)
0
c)
7
d)
4
|
Meera Rana answered |
Last digit of 794 is depends upon the the 94 i.e 1
Last digit of 875 is depends upon the the 75i.e 7
So the last digit of the whole expression will be 1 +7 = 8
So the last digit will be 8
Last digit of 875 is depends upon the the 75i.e 7
So the last digit of the whole expression will be 1 +7 = 8
So the last digit will be 8
What will be the last digit of 4356 x 567 x 4534- a)8
- b)5
- c)7
- d)4
Correct answer is option 'B'. Can you explain this answer?
What will be the last digit of 4356 x 567 x 4534
a)
8
b)
5
c)
7
d)
4
|
|
Arun Sharma answered |
Last digit of 4356 = 1 because cyclicity of 3 is 4 and 56 is completely
divisible by 4 so the last term of 34 is 1.
Last digit of 567 is 5 and last digit of 4534 is also 5
so that means the last digit of whole of the expression is 1 x 5 x 5 = 5 therefore the digit is 5.
divisible by 4 so the last term of 34 is 1.
Last digit of 567 is 5 and last digit of 4534 is also 5
so that means the last digit of whole of the expression is 1 x 5 x 5 = 5 therefore the digit is 5.
Find the maximum value of n such that 77! is per-fectly divisible by 720n.- a)35
- b)18
- c)17
- d)36
Correct answer is option 'C'. Can you explain this answer?
Find the maximum value of n such that 77! is per-fectly divisible by 720n.
a)
35
b)
18
c)
17
d)
36
|
|
Shalini Patel answered |
720 = 24 X 51 X 32
In 77! there would be 38 + 19 + 9 + 4 + 2 + 1 = 73 twos Æ hence [73/4] = 18 24s
In 77! there would be 25 + 8 + 2 = 35 threes -> hence [35/2] = 17 32s
In 77! there would be 15 + 3 = 18 fives
Since 17 is the least of these values, option (c) is correct.
Find the maximum value of n such that 570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81 is perfectly divisible by 30n.- a)12
- b)11
- c)14
- d)13
Correct answer is option 'B'. Can you explain this answer?
Find the maximum value of n such that 570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81 is perfectly divisible by 30n.
a)
12
b)
11
c)
14
d)
13
|
|
Shalini Patel answered |
Checking for the number of 2’s, 3’s and 5’s in the given expression you can see that the minimum is for the number of 3’s (there are 11 of them while there are 12 5’s and more than 11 2’s) Hence, option (b) is correct.
The remainder, when (1523 + 2323) is divided by 19, is:- a)4
- b)15
- c)0
- d)18
Correct answer is option 'C'. Can you explain this answer?
The remainder, when (1523 + 2323) is divided by 19, is:
a)
4
b)
15
c)
0
d)
18
|
|
Arun Sharma answered |
an +bn is always divisible by a + b when n is odd.
Therefore 1523 + 2323 is always divisible by 15 + 23 = 38.
As 38 is a multiple of 19, 1523 + 2323 is divisible by 19.
Therefore,the required remainder is 0.
Therefore 1523 + 2323 is always divisible by 15 + 23 = 38.
As 38 is a multiple of 19, 1523 + 2323 is divisible by 19.
Therefore,the required remainder is 0.
Find the remainder when 21875 is divided by 17.- a)8
- b)13
- c)16
- d)9
Correct answer is option 'B'. Can you explain this answer?
Find the remainder when 21875 is divided by 17.
a)
8
b)
13
c)
16
d)
9
|
|
Shubham Choudhary answered |
Solution:
We need to find the remainder when 21875 is divided by 17.
We can use the concept of modular arithmetic to solve this problem.
Modular arithmetic:
Modular arithmetic is a system of arithmetic for integers, which considers the remainder when one integer is divided by another integer. The remainder is called the modulus.
For example, 10 mod 3 = 1, which means that the remainder when 10 is divided by 3 is 1.
We can use the following property of modular arithmetic:
(a + b) mod n = [(a mod n) + (b mod n)] mod n
(a * b) mod n = [(a mod n) * (b mod n)] mod n
Using these properties, we can find the remainder when 21875 is divided by 17.
Solution:
21875 = 17 * 1285 + 10
The remainder when 21875 is divided by 17 is 10.
Therefore, the correct answer is option (B) 13.
Note: The answer given in the question is incorrect. The remainder when 21875 is divided by 17 is 10, not 13.
We need to find the remainder when 21875 is divided by 17.
We can use the concept of modular arithmetic to solve this problem.
Modular arithmetic:
Modular arithmetic is a system of arithmetic for integers, which considers the remainder when one integer is divided by another integer. The remainder is called the modulus.
For example, 10 mod 3 = 1, which means that the remainder when 10 is divided by 3 is 1.
We can use the following property of modular arithmetic:
(a + b) mod n = [(a mod n) + (b mod n)] mod n
(a * b) mod n = [(a mod n) * (b mod n)] mod n
Using these properties, we can find the remainder when 21875 is divided by 17.
Solution:
21875 = 17 * 1285 + 10
The remainder when 21875 is divided by 17 is 10.
Therefore, the correct answer is option (B) 13.
Note: The answer given in the question is incorrect. The remainder when 21875 is divided by 17 is 10, not 13.
What will be the unit digit when 4545- a)5
- b)0
- c)1
- d)9
Correct answer is option 'A'. Can you explain this answer?
What will be the unit digit when 4545
a)
5
b)
0
c)
1
d)
9
|
|
Aditya Kumar answered |
Here the last digit is depend upon the 5
The cyclicity of 5 is 1 so there is no need to divide the power with 1
because whatever is the power of 5 the last digit will remains 5 so the last
digit of 4545 will be 5 .
51 = 5
52 = 25
53 = 125
The cyclicity of 5 is 1 so there is no need to divide the power with 1
because whatever is the power of 5 the last digit will remains 5 so the last
digit of 4545 will be 5 .
51 = 5
52 = 25
53 = 125
Let N = 553 + 173 – 723. N is divisible by:- a)both 7 and 13
- b)both 3 and 13
- c)both 17 and 7
- d)both 3 and 17
Correct answer is option 'D'. Can you explain this answer?
Let N = 553 + 173 – 723. N is divisible by:
a)
both 7 and 13
b)
both 3 and 13
c)
both 17 and 7
d)
both 3 and 17
|
Vikas Kapoor answered |
We have N = 553 + 173 – 723 = (54 + 1)3 + (18 – 1)3 – 723
When N is divided by 3, we get remainders (1)3 + (- 1)3 – 0 = 0
Hence, the number N is divisible by 3.
Again N = (51 + 4)3 + 173 – (68 + 4)3
When N is divided by 17, the remainder is (4)3 + 0 – (4)3 = 0
Hence, the number is divisible by 17.
Hence, the number is divisible by both 3 and 17.
When N is divided by 3, we get remainders (1)3 + (- 1)3 – 0 = 0
Hence, the number N is divisible by 3.
Again N = (51 + 4)3 + 173 – (68 + 4)3
When N is divided by 17, the remainder is (4)3 + 0 – (4)3 = 0
Hence, the number is divisible by 17.
Hence, the number is divisible by both 3 and 17.
57 X 60 X 30 X 15625 X 4096 X 625 X 875 X 975- a)6
- b)16
- c)17
- d)15
Correct answer is option 'D'. Can you explain this answer?
57 X 60 X 30 X 15625 X 4096 X 625 X 875 X 975
a)
6
b)
16
c)
17
d)
15
|
|
Shalini Patel answered |
The given expression has fifteen 2’s and seventeen 5’s. The number of zeroes would be 15 as the number of 2’s is lower in this case. Option (d) is correct.
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