GATE 2018 EC GA Q.3: solution and answer

Official answer: BSource: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Divisibility rule for 3: a whole number is divisible by 3 exactly when the sum of its digits is divisible by 3. Where the missing digit sits in the number does not matter for this test. (The paper says "thousandths"; the digit is actually in the thousands place of 715■423.)

2. Add the known digits: $7+1+5+4+2+3=22$.

3. Call the missing digit $x$, where $x$ is one of $0,1,\dots ,9$. The digit sum is $22+x$, so $22+x$ must be a multiple of 3.

4. $22=3\times 7+1$, so $22$ leaves remainder 1 when divided by 3. For $22+x$ to leave remainder 0, $x$ must leave remainder 2, so $x\in \{2,5,8\}$.

5. The smallest of these is $x=2$. Check: the digit sum becomes $24=3\times 8$, so $7152423$ is divisible by 3. (Directly: $7152423=3\times 2384141$.)

6. Check the other options. (A) $x=0$: the digit sum is $22$, not a multiple of 3, so this fails. (C) $x=5$: the digit sum is $27$, which works, but $5>2$, so it is not the smallest. (D) $x=6$: the digit sum is $28$, not a multiple of 3, so this fails.

7. The smallest whole number that can go in place of ■ is $2$, which is option (B).