This is a common Quantitative Aptitude (Number System) interview and practice question. Below is the full answer with a clear, step-by-step explanation.

Question

Find the smallest number which when divided by 8, 12, and 16 leaves a remainder of 5 in each case.

Answer & Explanation

Answer: 53

Why 53 is the correct number

When a number N leaves the same remainder r when divided by several divisors, we can write

N = k·(LCM of the divisors) + r

where k is a non‑negative integer and the LCM (least common multiple) is the smallest number that is a multiple of every divisor.

In this problem the divisors are 8, 12 and 16.

Find the LCM
- Prime factorisations:
  8 = 2³ 12 = 2²·3 16 = 2⁴
- Take the highest power of each prime that appears:
  2⁴ (from 16) and 3¹ (from 12).
- LCM = 2⁴·3 = 16·3 = 48.
Apply the remainder condition
The required remainder is 5, so any number that works must be of the form

N = 48·k + 5

with k ≥ 0 and N > 5 (otherwise the remainder would be the number itself).

Choose the smallest k
The smallest non‑negative integer k that makes N larger than the remainder is k = 1.

N = 48·1 + 5 = 53.

Check
- 53 ÷ 8 = 6 remainder 5 (8·6 = 48, 53‑48 = 5)
- 53 ÷ 12 = 4 remainder 5 (12·4 = 48, 53‑48 = 5)
- 53 ÷ 16 = 3 remainder 5 (16·3 = 48, 53‑48 = 5)

All three give remainder 5, so 53 satisfies the condition and is the smallest such number.

**

Quick Method

(exam‑time shortcut)**

Compute the LCM of the divisors – just look for the highest power of each prime.
- 8 = 2³, 12 = 2²·3, 16 = 2⁴ → LCM = 2⁴·3 = 48.
Add the required remainder – because the number must be “LCM + remainder”.
- 48 + 5 = 53.

That’s it. No need to test larger multiples; the first multiple (k = 1) already gives the smallest possible answer. If the remainder were larger than the LCM, you would increase k, but here 5 < 48, so the answer is simply LCM + remainder.