TCS Ninja Divisibility Questions Part 1

Video solution:

Video solution:

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
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:

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:

When, 6 divides 1201, 1202, 1203, 1205, 1207 respectively; it leaves remainder 1,3,5,1.

The product of these remainder = 15

When, 15 is divided by 6, remainder is 3.

Video solution:

Video solution:

Of the given alternatives,

Remainder = 9 because, when it is subtracted from the number, the unit's digit will be 5 and number is divisible by 45