2024 CSAT — Q18

If a divisor $d$ leaves remainder $1$ on both $421$ and $427$, then $d$ divides both $421-1 = 420$ and $427-1 = 426$. Hence

Compute the gcd by subtraction (fast under the clock): $426 - 420 = 6$, and $\gcd(420,6) = 6$. So $d$ is a divisor of $6$: $\{1, 2, 3, 6\}$.

The load-bearing constraint: a divisor that leaves remainder $1$ must be strictly greater than $1$ (you cannot have a remainder of $1$ when dividing by $1$). Drop $d=1$.

Valid divisors: $\{2, 3, 6\}$ — three of them. (Quick check: $421 = 6\cdot70+1$, $427 = 6\cdot71+1$, etc., all give remainder $1$.)

Answer: (c) 3.

“Same remainder on two numbers” is a gcd-of-the-differences template, every single time. The key fact volume-players skip: subtract the common remainder first, then the divisors are exactly the divisors of the gcd that exceed the remainder. Here the remainder is $1$, so we keep divisors $> 1$. Had the remainder been $2$, we’d keep divisors $> 2$ (only $3$ and $6$, giving $2$). That single guard — divisor must exceed the remainder — is the whole difference between (c) and (d), and it is the most common silent error on this template across years. Sanity check: you can ignore the original numbers entirely once you have $\gcd(420,426)=6$; the problem is now pure divisor-counting.