SSC CHSL 2022 PYQ — Question 63

SSC CHSL - 2 June 2022 - Shift 2 • Quantitative Aptitude • Q63

Find the smallest four-digit number which is divisible by 4, 8, 12, and 16.

Show answer & explanation

Correct answer: B

Explanation

The correct answer is Option 2 i.e. 1008.

The smallest 4-digit number is 1,000

Now, LCM (4, 8, 12, 16) = 48

On dividing 1000 by 48, the remainder = 40

Now, the smallest 4-digit number divisible by 4, 8, 12, 16 is = 1,000 - 40 + 48 = 1,008

Hence, the required number is 1,008