TCS Divisibility Quiz-1 Questions and Answers

n divided by 5 yields a remainder equal to 3 is written as follows

n = 5 k + 3 , where k is an integer.

add 2 to both sides of the above equation to obtain

n + 2 = 5 k + 5 = 5(k + 1)

The above suggests that n + 2 divided by 5 yields a remainder equal to zero.

Video solution:

If n is divisible by 3, 5 and 12 it must a multiple of the lcm of 3, 5 and
12 which is 60.

n = 60 k

n + 60 is also divisible by 60 since

n + 60 = 60 k + 60 = 60(k + 1)

Video solution:

One may answer this question using a calculator and test for divisibility by 3.
However we can also test for divisibilty by adding the digits and if
the result is divisible by3 then the number is divisible by 3.

3 + 3 + 9 = 15 , divisible by 3.

3 + 4 + 2 = 9 , divisible by 3.

5 + 5 + 2 = 12 , divisible by 3.

1 + 1 + 1 + 1 = 4 , not divisible by 3.

The number 1111 is not divisible by 3.

Video solution:

Let xy be the whole number with x and y the two digits that make
up the number. The number is divisible by 3 may be written as follows

10 x + y = 3 k

The sum of x and y is equal to 9.

x + y = 9

Solve the above equation for y

y = 9 - x Substitute y = 9 - x in the equation 10 x + y = 3 k to obtain.

10 x + 9 - x = 3 k

Solve for x

x = (k - 3) / 3

x is a positive integer smaller than 10

Let k = 1, 2, 3, ... and select the first value that gives x as an integer. k = 6 gives x = 1

Find y using the equation y = 9 - x = 8

The number we are looking for is 18 and the answer is A.
It is divisible by 3 and the sum of its digits is equal to 9 and it
is the smallest and positive whole number with such properties.

Video solution:

We first expand (2n + 2)²

(2n + 2)²= 4n² + 8n + 4

Factor 4 out.

= 4(n² + 2n + 1)

(2n + 2)² is divisible by 4 and the remainder is equal to 0.


Video solution:

Given N = 80pq2pq is exactly divisible by 120.
Since the last digit of the dividend is 0, the last digit of N also should be 0.
Then obviously, q = 0.
i.e., N = 80p02p0

Now we have to find p.
We know 120 = 3 x 5 x 8 and 3, 5 and 8 are co-primes.
If N is divisible by 120 then it is divisible by 3, 5 and 8.
We know that, \"If a number is divisible by 3 then the sum of its digits
also divisible by 3\"
Then, 8 + 0 + p + 0 + 2 + p + 0 = 10 + 2p
ie., 10 + 2p must be divisible by 3

Then the possible values of p are 1,4,7,10,13 and so on
Since p is a digit, the possible values are 1, 4, 7

Now, N must be divisible by 8 also.
We know that, \"If a number is divisible by 8 then its last 3 digits
also divisible by 8\"
Here 2p0 is a multiple of 8.

Now put all the above possible values of p then we have 2p0 = 210 or 240 or 270
From these, 240 is a multiple of 8.
Then the value of p = 4.
and N = 8040240
Hence the required sum = 8 + 0 + 4 + 0 + 2 + 4 + 0 = 18

Video solution:

Given that, 53p26p3 is a 7 digit number divisible by 9.
We know that, if a number is divisible by 9 then the sum of digits
s also divisible by 9.
Then 5 + 3 + p + 2 + 6 + p + 3 = 19 + 2p is divisible by 9
If p = 1 then 19 + 2p = 21 which is not divisible by 9.
If p = 2 then 19 + 2p = 24 which is not divisible by 9.

Proceeding like this,
we would have 19 + 2p divisible by 9 when p = 4

Given that, the 2nd number 757qp = 757q4 is divisible by 8.
Then the last 3 digits 7q4 is divisible by 8.
Then the possible values are 704, 714, 724, 734, 744, 754, 764, 774, 784, 794.

From these, 704, 744 and 784 are the multiples of 8.
Then the possible value for q are 0,4 and 8.
We have to find the minimum value of p + q
p + q = 4 + 0 = 4,
or 4 + 4 = 8 or 4 + 8 = 12
Hence the required answer is 4.

Video solution:

If you don\'t know the above rule, this problem is really calculation intensive.
But by applying the above rule, when 1201, 1203, 1205, 1207 divided by 6,
leaves remainders 1, 3, 5, 1. The product of these remainders = 15.
When 15 is divided by 6, Remainder is 3.

Video solution:

By applying rule 2, we divide the terms of the above expression individually,
and add them to get the final remainder. But from 4! onwards
all the terms leave a remainder 0 when divided by 24.
So the remainder = 1 + 2 + 6 + 0 + 0....... = 9

Video solution:

Divisibility for 45 is, the given number should be divisible by 9
as well as with 5.
Let the given number is N.
N has unit digit 5 so it is divisible by 5.
Now the divisibility for 9 is sum of the digits of N should be
divisible by 9.
Digit sum of N = (1 + 2 + 3 + ... + 9) + 1 × 10 + (1 + 2 + 3 + ... + 9) + 2 × 10 +
(1 + 2 + 3 + ... + 9) + 3 × 10 + (1 + 2 + 3 + ... + 9) + 4 × 6 +
(1 + 2 + 3 + 4 + 5)
= 45 + 10 + 45 + 20 + 45 + 30 + 45 + 24 + 15 = 279
279 when divided by 9, remainder 0. So N can be written as 9m
So, N = 5k = 9m
So the given N is both multiple of 5 and 9. So it is exactly divisible by 45. So remainder = 0

Video solution: