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MAT Exam 2022 Paper

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Question : 89 of 200

Marks: +1, -0

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?

2519
5039
1079
979

Solution:

Concept:

The given remainders are each exactly

1

less than the corresponding divisors.

Explanation:

Let the required number be

N

.

When

N

is divided by

5

, the remainder is

4

, so

N+1

is divisible by

5

.

Similarly,

N+1

is divisible by

6

,

7

,

8

, and

9

.

Therefore,

N+1

must be the least common multiple of

5, 6, 7, 8,

and

9

.

\mathrm{LCM}(5,6,7,8,9) = 2520

So,

N+1 = 2520

, which gives

N = 2520 - 1 = 2519

.

Answer:

The smallest such number is

2519

, i.e., Option A.

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