CSAT Solved Papers/ 2023/Q25

2023 CSAT — Q25

Quant Number theory 2.5 marks Medium

Let $x$ be a positive integer such that $7x + 96$ is divisible by $x$ . How many values of $x$ are possible?

A 10
B 11
C 12 Answer
D Infinitely many

Worked rationale

If $x \mid 7x + 96$ , then since $x \mid 7x$ , we must have $x \mid 96$ . Conversely every divisor of $96$ works. So the number of valid $x$ equals the number of positive divisors of $96$ .

Factorise: $96 = 2^5 \times 3$ . The divisor count is

(5 + 1)(1 + 1) = 6 \times 2 = 12.

Answer: (c) 12.

Why the other options miss

A

mishandled a repeated factor: mis-factors $96$ (e.g. $2^4 \cdot 3 \cdot ?$ ) or applies the exponent $+1$ rule to a wrong exponent, undercounting divisors.
B

off by one: computes $12$ then drops $x = 1$ or $x = 96$ as “trivial,” or forgets to add $1$ to one exponent.
D

solved the wrong question: reasons ” $7x$ is always divisible by $x$ , so any $x$ works,” forgetting the $+96$ term that pins $x \mid 96$ .

Specialist insight

The whole problem is the one-line reduction $x \mid 7x + 96 \iff x \mid 96$ (subtract the obvious multiple $7x$ ). Then it is a pure divisor-count: $96 = 2^5\cdot 3 \Rightarrow (5+1)(1+1) = 12$ . The seductive wrong answer is (d) “infinitely many,” which catches anyone who sees only the $7x$ and misses that the constant $96$ bounds the divisors. CSAT routinely hides a divisor-count behind a divisibility phrasing.

The trap, in one line

$x\mid 7x+96 \iff x\mid 96$ ; $96 = 2^5\cdot 3$ has $(5+1)(1+1)=12$ divisors $\Rightarrow$ (c).

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