number systems Model Questions & Answers, Practice Test for ibps clerk mains

Answer: (b)

Since 111111 is divisible by each one of 7, 11 and 13, so each one of given type of numbers is divisible by each one of 7, 11, and 13.

as we may write, 222222 = 2 × 111111, 333333 = 3 × 111111, etc.

Answer: (a)

$3^521= 3^130 × 4 + 1$

As we know $3^1$ = 3 will leave remainder = 3 when divided by 8

Answer: (a)

$a^2 – b^2$ = 35

⇒ (a + b) (a – b)

= 35 ×1 or 7 × 5

Hence, two such pairs are (1,35) and (5,7).

Answer: (b)

Let the common remainder be x.

Then numbers (34041 – x) and (32506 – x) would be completely divisible by n.

Hence the difference of the numbers (34041 – x) and (32506 – x) will also be divisible by n

or (34041 – x – 32506 + x) = 1535 will also be divisible by n.

Now, using options we find that 1535 is divisible by 307.

Answer: (c)

Required numbers are of the form of 12q – 2

i.e., 10, 22, 34, 46, 58, 70, 82, 94

∴ Total sum = 10 + 22 + 34 + 46 + 58 + 70 + 82 + 94 = 416