2023 CSAT — Q69

Statement 1 — fewest coins whose weights sum to $78$ gm. Look mod $5$: only the $2$-gm coin is non-zero mod $5$ ($2,5,10,25,50 \equiv 2,0,0,0,0$). We need $\sum \equiv 78 \equiv 3 \pmod 5$, so the number $k$ of $2$-gm coins satisfies $2k \equiv 3 \pmod 5 \Rightarrow k \equiv 4 \pmod 5$, hence $k \ge 4$.

Take $k = 4$ (the four $2$-gm coins give $8$ gm); the remaining $78 - 8 = 70$ gm from $\{5,10,25,50\}$ needs at least $3$ coins ($50 + 10 + 10 = 70$). Total $4 + 3 = 7$ coins, and no combination does it in fewer. So at least $7$ coins — Statement 1 is true.

Statement 2 — weigh $78$ gm on a balance, where coins may sit on either pan (so subtraction is allowed). Then

using only $4$ coins $< 7$. So less than $7$ coins suffices — Statement 2 is true.

Answer: (c) Both 1 and 2.

The two statements use different operations: “buy/sum to $78$” (coins add only) vs “weigh $78$” (balance — coins add or subtract). For the sum, the mod-$5$ argument is decisive: $2k\equiv 3 \pmod 5$ forces $k\ge 4$ two-gm coins, pushing the minimum to $7$. For the weighing, two-pan placement gives $78 = 50+25+5-2$ in $4$ coins. Reading “buy” vs “weigh” precisely — and knowing a balance allows subtraction — is exactly what separates (c) from the traps.