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SSC CHSL 2019 PYQ — Question 51

SSC CHSL - 11th July 2019 - Shift 1 • Quantitative Aptitude • Q51

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The ten digit number 2x600000y8 is exactly divisible by 24. If x ≠ 0 and y ≠ 0, then the least value of (x + y) is equal to:

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Correct answer: C

Explanation

Correct Answer is option 3 i.e. 5

Concept

Understanding

Application

If a number is divisible by 24, the number will also be divisible by 8 and 3.

For a number to be divisible by 8, the last three digits should be divisible by 8.

For a number to be divisible by 3, the sum of the digits should be divisible by 3.

Given that the number 2x600000y8 is exactly divisible by 24.

Here,

For, y = 4, the last three digits, i.e. 048 is divisible by 8.

For divisibility by 3:

20 + x should be divisible by 3

Then, x = 1

So, we have:

x = 1 and y = 4

Now,

x + y = 1 + 4 = 5