GATE 2023 EC GA Q.3: solution and answer

Official answer: CSource: Official answer key, GATE 2023 EC (IIT Kanpur)

1. Check the options against the "distinct digits" condition first. (A) $123555789$ uses the digit 5 three times and (D) $99999$ uses 9 five times, so both are ruled out even though their digits do add up to 45.

2. Find the fewest digits that can work. The largest sum you can get from 8 distinct digits is $9+8+7+6+5+4+3+2=44<45$. So no number with 8 or fewer distinct digits reaches 45, and the number needs at least 9 digits.

3. The ten digits add up to $0+1+2+\cdots +9=\frac{9\times 10}{2}=45$. If 9 distinct digits add up to 45, the one digit left out must be $45-45=0$. So the digits are exactly $1,2,\dots ,9$, each used once.

4. Using 0 as a tenth digit keeps the sum at 45, but the number then has 10 digits, and every 10-digit number is larger than every 9-digit number. So the smallest number has exactly 9 digits.

5. Once the number of digits is fixed, the smallest arrangement puts the digits in ascending order, so the smallest digit sits in the highest place value. That gives $123456789$.

6. Check (B): $123457869$ uses the same nine digits. It agrees with $123456789$ in the first five digits, but its sixth digit is 7 where (C) has 6, so (B) is larger. The smallest such number is $123456789$, which is option (C).