number system Model Questions & Answers, Practice Test for ibps clerk prelims
If n is a whole number greater than 1, then $n^2 (n^2 – 1)$ is always divisible by
Answer: (c)
Given that n is greater then 1, then $n^2 (n^2 – 1)$ is always divisible by 12.
Example 1: Put n = 2, then
$n^2 (n^2 – 1) = (2)^2 (2^2 – 1)$
= 4 × 3 = 12.
Hence $n^2 (n^2 – 1)$ is divisible by 12 but not by 24, 48 and 60 for n = 2.
A two-digit number is 9 more than four times of the number obtained by interchanging its digits. If the product of digits in the two-digit number is 8, then what is the number?
Answer: (c)
Let two digits of the number are x and y, then
xy = 8 ⇒ y = 8/x ...(i)
(10x + y) = 4 (10y + x) + 9
10x + y = 40y + 4x + 9
6x = 39y + 9
2x = 13y + 3
$2x = 104/x + 3$ $(∵ y = 8/x)$
$2x^2 – 3x – 104 = 0$
⇒ x = 8, $13/2$
∴ x = 8, y = 1
Required number = 81
$12^55/3^11 + 8^48/16^18$ will give the digit at units place as
Answer: (d)
$12^55/3^11 = 3^44.{4}^55$ → 4 as units place.
Similarly, $8^48/16^18 = 2^72$ → 6 as the units place.
Hence, 0 is the answer.
How many numbers are there in the set S = {200, 201, 202, ...,800} which are divisible by neither of 5 or 7?
Answer: (c)
Total numbers in the set = (800 – 200) + 1 = 601
Number of numbers which are divisible by 5
= $(800 - 200)/5 + 1 = 121$
Number of numbers which are divisible by 7
= $(798 - 203)/7 + 1= 86$
Number of numbers which are divisible by both 5 & 7
= $(770 - 210)/35 + 1 = 17$
∴ Number of numbers which are either divisible by 5 or 7 or both
= (121 + 86) – 17 = 190
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