Number System Questions for IBPS, SBI PO, SO, Clerk Exams - 1

Ans: B

Given expression = (397)² + (104)² + 2 × 397 × 104

= (397 + 104)² = (501)² = (500 + 1)²

= (500)² + 1² + 2 x 500 x 1 = 250000 + 1 + 1000 = 251001.

Ans: D

Let N = (1 x 3 x 5 x 7 x ….. x 99) x 128.

Clearly, N contains 10 multiplies of 5 (5, 15, 25, 35, 95) and only one multiple of 2 i.e., 128 or 2 .

Clearly, highest power of 5 in N is greater than that of 2.

∴ Number of zeros in N = Higher power of 2 in N = 7.

Ans: D

Sum of all digits = 12 which is divisible by 3.

So, the given number is divisible by 3.

(Sum of digits at odd places) - (Sum of digits at even places) = 6 – 6 = 0.

So, the given number is divisible by 11.

Ans: A

Consider the number 17325.

Its unit digit is 5. So, it is divisible by 5.

Sum of its digits = (5 + 2 + 3 + 7 + 1) = 18, which is divisible by 3.

So, the given number is divisible by 3.

And, since 5 and 3 are co-primes.

So the given number is divisible by (5 × 3) i.e., 15.

Ans: A

We know that (xⁿ - aⁿ) is always divisible by (x - a) for all values of n.