CSAT Solved Papers/ 2022/Q65

2022 CSAT — Q65

Quant Number theory 2.5 marks Medium

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 $3$ ?

A 1063
B 1073
C 1083 Answer
D 1183

Worked rationale

” $N$ leaves remainder $3$ when divided by each of $6,9,12,15,18$ ” means $N - 3$ is a common multiple of all five, i.e. a multiple of their LCM.

\operatorname{lcm}(6,9,12,15,18): \quad 6=2\cdot 3,\ 9=3^2,\ 12=2^2\cdot 3,\ 15=3\cdot 5,\ 18=2\cdot 3^2 \;\Rightarrow\; 2^2\cdot 3^2\cdot 5 = 180.

So $N = 180k + 3$ . We need $N > 1000$ : $180k + 3 > 1000 \Rightarrow 180k > 997 \Rightarrow k \ge 6$ (since $180\times 5 = 900 < 997 < 1080 = 180\times 6$ ). Take $k = 6$ :

N = 180 \times 6 + 3 = 1083.

Answer: (c) 1083.

Why the other options miss

A

an arithmetic slip: adds $3$ to a non-multiple of $180$ ( $1060$ ) or uses a wrong LCM.
B

reached for the wrong LCM: uses an LCM around $70$ or mis-adds the remainder.
D

overshot the smallest: takes $k = 7$ ( $180\times 7 + 3$ ), one multiple past the smallest above $1000$ .

Specialist insight

The “same remainder” structure collapses to LCM-plus-offset: $N = \operatorname{lcm}(\dots)\cdot k + \text{remainder}$ . Get the LCM right ( $180$ , not the product), then pick the smallest $k$ clearing $1000$ ( $k=6$ , since $900+3 = 903 \le 1000$ ). The trap (d) is the next multiple up — answer the smallest, not just any valid one.

The trap, in one line

$N = 180k + 3$ ; smallest with $N > 1000$ is $k=6 \Rightarrow 1083 =$ (c).

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