5 CAT Remainders questions you should solve

Answer: (d)

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

Answers: (c)

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.

Answers: (b)

First of all, we have to identify such 2 digit numbers.

Obviously, they are 10, 17, 24, ….94

The required sum = 10 + 17 … 94.

Now this is an A.P. with a = 10, n = 13 and d = 7

Hence, the sum is

Answers: (d)

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.

Answers: (d)

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.