CSAT Solved Papers/ 2025/Q47

2025 CSAT — Q47

Quant Number theory 2.5 marks Medium

Let $P = QQQ$ be a $3$ -digit number. What is the HCF of $P$ and $481$ ?

A 1
B 13
C 37 Answer
D 481

Worked rationale

A repdigit $QQQ$ factors as

QQQ = Q \times 111 = Q \times 3 \times 37,

where $Q$ is a single nonzero digit ( $1 \le Q \le 9$ ). And $481 = 13 \times 37$ .

\gcd(QQQ,\,481) = \gcd(3 \cdot 37 \cdot Q,\ 13 \cdot 37) = 37 \times \gcd(3Q,\ 13).

Now $13$ is prime and $13 \mid 3Q$ would require $13 \mid Q$ — impossible for a digit $Q \le 9$ . So $\gcd(3Q, 13) = 1$ , and the HCF is exactly $37$ , independent of which digit $Q$ is.

Answer: (c) 37.

Why the other options miss

A

missed a case: misses that $111 = 3 \times 37$ shares the factor $37$ with $481$ , concluding the numbers are coprime.
B

wrong formula: factors $481 = 13 \times 37$ but picks the wrong shared prime ( $13$ ), not checking that $13 \nmid QQQ$ .
D

solved the wrong question: assumes $481 \mid QQQ$ for some $Q$ (it never does, since $13 \nmid QQQ$ ), treating $481$ as a common divisor.

Specialist insight

The unlock is the repdigit identity $\overline{QQQ} = 111\,Q = 3\cdot 37\cdot Q$ — $111$ ‘s factor $37$ is the bridge to $481 = 13\times 37$ . Once both numbers are in prime form the gcd is just the shared primes: $37$ is always common; $13$ never is (it cannot divide a single digit). The beautiful part is that the answer does not depend on $Q$ — true for $111, 222, \dots, 999$ alike — which is exactly why the question can pose a “variable” $P$ and still have a unique answer. Memorising $111 = 3\cdot37$ , $1001 = 7\cdot11\cdot13$ , $1111 = 11\cdot101$ pays off repeatedly on CSAT structured-number items.

The trap, in one line

$QQQ = 3\cdot37\cdot Q$ and $481 = 13\cdot37$ share only $37$ ( $13$ can't divide a single digit) — HCF $= 37$ for every digit $Q$ .

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