CSAT Solved Papers/ 2024/Q54

2024 CSAT — Q54

Quant Number theory 2.5 marks Easy

$32^5 + 2^{27}$ is divisible by

A 3
B 7
C 10 Answer
D 11

Worked rationale

Rewrite everything to a common base $2$ . Since $32 = 2^5$ , we have $32^5 = 2^{25}$ . Then

32^5 + 2^{27} = 2^{25} + 2^{27} = 2^{25}\big(1 + 2^2\big) = 2^{25} \times 5.

So the number is $2^{25}\cdot 5$ . Its prime factors are only $2$ and $5$ . Among the options, only $10 = 2\times5$ divides it (both prime factors present); $3$ , $7$ , $11$ do not.

Answer: (c) 10.

Why the other options miss

A

solved the wrong question: guesses divisibility by $3$ without factoring; $2^{25}\cdot5$ has no factor of $3$ .
B

mishandled a repeated factor: mishandles the exponents (e.g. reads $32^5 = 2^{10}$ ), producing a spurious factor.
D

wrong formula: tries a $a^n + b^n$ divisibility rule meant for odd $n$ with distinct bases, which does not apply to a single base.

Specialist insight

The entire item collapses once you put both terms on the same base: $32 = 2^5$ , so the sum is $2^{25}(1 + 4) = 2^{25}\cdot5$ . Factoring out the smaller power exposes the complete prime factorisation, and divisibility becomes a one-glance check — only $2$ and $5$ are present, so $10$ works and nothing else on the list does. The discipline that scores: convert to a common base, factor out the lowest power, read the factors. Never test divisibility by trial division on a giant power.

The trap, in one line

$32^5 = 2^{25}$ , so the sum is $2^{25}(1+4) = 2^{25}\cdot5$ — prime factors only $2$ and $5$ , hence divisible by $10$ alone.

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