Prelims 2022 · CSAT / Quantitative Aptitude · Question 65
What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?
Answer
1083
We need a number N > 1000 such that dividing by 6, 9, 12, 15, 18 leaves remainder 3. So N - 3 must be divisible by all these numbers.
- Key fact: (\text{LCM}(6,9,12,15,18)=180). Hence (N = 180k + 3).
- Smallest such number greater than 1000: (180 \times 5 + 3 = 903) (too small), (180 \times 6 + 3 = 1083).
Option check:
- (a) 1063: (1063-3=1060), not divisible by 9, 12, 15, 18. Incorrect.
- (b) 1073: (1073-3=1070), not divisible by 6, 9, 12, 15, 18. Incorrect.
- (c) 1083: (1083-3=1080), divisible by 6, 9, 12, 15, 18. Correct.
- (d) 1183: (1183-3=1180), not divisible by 9, 12, 15, 18. Incorrect.