2025 CSAT — Q18

Translate to congruences and drop the redundant moduli first. $N \equiv 7 \pmod 9$ already forces $N \equiv 1 \pmod 3$ (since $7 \equiv 1 \pmod 3$), so the mod-$3$ condition is free. The mod-$6$ condition $N \equiv 4 \pmod 6$ means $N$ is even and $N \equiv 1 \pmod 3$ (consistent). So the binding conditions are:

Combine mod $9$ and mod $5$ (CRT, lcm $45$): solve $N \equiv 7 \pmod 9$ and $N\equiv 3\pmod5$. Listing $7,16,25,34,43,\dots$, the value $43$ satisfies $43 \equiv 3 \pmod 5$. So $N \equiv 43 \pmod{45}$.

Now impose even: $43$ is odd, so add $45$ to get $88$ (even); the joint period is $\operatorname{lcm}(45,2) = 90$, giving $N \equiv 88 \pmod{90}$. (Check mod $6$: $88 = 14\cdot6

• 4$ ✓.)

Smallest $4$-digit: $N = 90k + 88 \ge 1000 \Rightarrow k \ge 10.13 \Rightarrow k = 11$:

Verify: $1078 = 3\cdot359+1,\ 5\cdot215+3,\ 6\cdot179+4,\ 9\cdot119+7$ — all four ✓.

Answer: (c) 1078.

The exam-fast move is culling redundant moduli before any CRT work: mod $9$ swallows mod $3$, and mod $6$ splits into “even” plus a mod-$3$ fact already covered — so four conditions collapse to two congruences and a parity. Notice too the common deficiency: each remainder is exactly $2$ short of the divisor ($3{-}1,\ 5{-}3,\ 6{-}4,\ 9{-}7 = 2$), so $N+2$ is divisible by all of $3,5,6,9$, i.e. $N+2 \equiv 0 \pmod{\operatorname{lcm}=90}$, giving $N \equiv -2 \equiv 88 \pmod{90}$ in one line. Spotting the uniform deficiency is the elegant shortcut that beats brute CRT and is exactly the structural eye CSAT rewards.