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Question

Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

CAT 2021

A
2519
CORRECT ANSWER WRONG ANSWER
B
5039
CORRECT ANSWER WRONG ANSWER
C
1079
CORRECT ANSWER WRONG ANSWER
D
979
CORRECT ANSWER WRONG ANSWER
Solution

Correct option is

(A)

Explanatory Answer :

As we know , When a number is divided by 8, a remainder of 7 can be thought of as a remainder of -1. So,

N = 5a - 1 or N + 1 = 5a

N = 6b - 1 or N + 1 = 6b

N = 7c - 1 or N + 1 = 7c

N = 8d - 1 or N + 1 = 8d

N = 9e - 1 or N + 1 = 9e,

N + 1 can be expressed as a multiple of (5, 6, 7, 8, 9)

N + 1 = 5a * 6b * 7c * 8d * 9e

or N = (5a×6b×7c×8d×9e) - 1

Smallest value of N will be when we find the smallest common multiple of (5, 6, 7, 8, 9)

or LCM of (5, 6, 7, 8, 9)

N = LCM (5, 6, 7, 8, 9) - 1

= 2520 - 1

= 2519.

Hence, A is the correct option.