CSAT Solved Papers/ 2022/Q58

2022 CSAT — Q58

Quant Number theory 2.5 marks Hard

What is the remainder when $91 \times 92 \times 93 \times 94 \times 95 \times 96 \times 97 \times 98 \times 99$ is divided by $1261$ ?

A 3
B 2
C 1
D 0 Answer

Worked rationale

Factor the divisor: $1261 = 13 \times 97$ (both prime).

Look for these factors inside the product $91 \times 92 \times \cdots \times 99$ :

$97$ is one of the factors directly.
$91 = 13 \times 7$ , so $13$ divides the product too.

Since $13$ and $97$ are distinct primes both dividing the product, their product $1261$ divides it:

1261 = 13 \times 97 \mid (91 \times 92 \times \cdots \times 99).

So the remainder is $0$ .

Answer: (d) 0.

Why the other options miss

A

skipped the factorisation: all assume a non-trivial remainder because the product “looks like it won’t divide,” never factoring the divisor $1261 = 13\times 97$ to see that both prime factors ( $97$ directly, $13$ inside $91$ ) actually sit inside the product.

Specialist insight

Never brute-force a nine-term product. Factor the divisor ( $1261 = 13 \times 97$ ) and check whether each prime appears among the consecutive factors $91$ – $99$ : $97$ is present, and $91 = 7\times 13$ supplies the $13$ . Distinct primes both present $\Rightarrow$ the product is a multiple of $1261$ , remainder $0$ . The ” $91$ hides a $13$ ” observation is the whole game.

The trap, in one line

$1261 = 13 \times 97$ ; the run $91$ – $99$ contains $97$ and $91 = 13\times 7$ , so it divides exactly $\Rightarrow$ remainder $0 =$ (d).

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