CSAT Solved Papers/ 2023/Q45

2023 CSAT — Q45

Quant Number theory 2.5 marks Medium

How many natural numbers are there which give a remainder of $31$ when $1186$ is divided by these natural numbers?

A 6
B 7
C 8
D 9 Answer

Worked rationale

If dividing $1186$ by $d$ leaves remainder $31$ , then $d \mid (1186 - 31)$ and $d > 31$ (a divisor must exceed its remainder).

1186 - 31 = 1155 = 3 \times 5 \times 7 \times 11.

All divisors of $1155$ (there are $2^4 = 16$ ): $1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155$ .

Keep only those greater than $31$ :

33, 35, 55, 77, 105, 165, 231, 385, 1155 \quad\text{— that is } 9 \text{ values.}

Answer: (d) 9.

Why the other options miss

A

a remainder-rule slip: applies the $d > 31$ cut too aggressively (e.g. demands $d > 35$ or drops valid divisors like $33, 35$ ), undercounting.
B

off by one: drops two boundary divisors (commonly $33$ and $35$ , the ones just above $31$ ).
C

off by one: omits exactly one qualifying divisor (often $1155$ itself, “too big,” or $33$ ).

Specialist insight

Two non-negotiable rules: (1) $d \mid (1186 - 31)$ , and (2) $d > 31$ — the divisor must be larger than the remainder it produces, or the remainder $31$ is impossible. Many candidates nail the factorisation $1155 = 3\cdot 5\cdot 7\cdot 11$ (16 divisors) but forget rule (2) and answer $16$ , or apply it sloppily. List the divisors, then keep the nine that exceed $31$ .

The trap, in one line

$d\mid 1186-31 = 1155 = 3\cdot 5\cdot 7\cdot 11$ and $d > 31$ ; nine divisors exceed $31 \Rightarrow$ (d).

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