2021 CSAT — Q36

Take three distinct non-zero digits $a,b,c$ that are multiples of $3$ (so each $\in\{3,6,9\}$, and in fact $\{a,b,c\}=\{3,6,9\}$). Form all $3!=6$ three-digit numbers with no repetition. Each digit lands in each place exactly $2$ times, so

Here $a+b+c = 3+6+9 = 18$, giving $S = 222\times 18 = 3996$.

Factor $222 = 2\times 3\times 37$, and note $222 = 3\times 74$.

• Statement 1: $S = 222(a+b+c) = 74\cdot 3(a+b+c)$, divisible by $74$. ($3996 = 74\times 54$.) Correct.
• Statement 2: $222$ already contributes one $3$, and the digit sum $a+b+c=18$ contributes another factor of $3$ (indeed $9$), so $S$ is divisible by $9$. ($3996 = 9\times 444$.) Correct.

Answer: (c) Both 1 and 2.

The engine is the identity $S = 222\,(a+b+c)$ — each digit occupies each of the three places exactly $(n-1)! = 2$ times. Everything then follows from factoring $222 = 2\cdot 3\cdot 37 = 3\cdot 74$ and from $a+b+c=18$ contributing $3^2$. The deep point both wrong answers miss is where the factors of $3$ live: one in $222$, one (actually two) in the digit sum — together they over-satisfy the divisibility by $9$. Pin the sum-of-all-permutations formula first; the divisibility claims are then pure factoring.