What is the least number which, when divided by 4, 6, 10 and 15, leaves a remainder of 3 in each case, but is exactly divisible by 11?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 363
A number leaving remainder 3 on division by each of them is (a multiple of their LCM) \(+\,3\). Try the multiples in turn until the number is divisible by 11.
LCM \(=60\), so the number is \(60 k+3\)
\(k=1\): 63, not divisible by 11
\(k=2\): 123, not divisible by 11
\(k=3\): 183, not divisible by 11
\(k=4\): 243, not divisible by 11
\(k=5\): 303, not divisible by 11
\(k=6\): 363, divisible by 11
The least such number is 363.
Why the other options are incorrect
Option (2) 183 — Incorrect. 183 leaves remainder 3 on each division, but it is not divisible by 11.
Option (3) 723 — Incorrect. 723 leaves remainder 3 on each division, but it is not divisible by 11.
Option (4) 63 — Incorrect. 63 leaves remainder 3 on each division, but it is not divisible by 11.