UPSC CDS 1 2020 Math Paper

Concept:Express the 3‑digit number in expanded form and add its cyclic permutations to factor out a common multiple.Explanation:Let $XYZ$ be a 3‑digit number with digits $X$, $Y$, $Z$ where $X \neq 0$.Expanded form: $XYZ = 100X + 10Y + Z$.Similarly, $YZX = 100Y + 10Z + X$ and $ZXY = 100Z + 10X + Y$.Add them: $S = (100X+10Y+Z) + (100Y+10Z+X) + (100Z+10X+Y)$.Group like terms: $S = (100X+X+10X) + (10Y+100Y+Y) + (Z+10Z+100Z)$.Simplify: $S = 111X + 111Y + 111Z = 111(X+Y+Z)$.Now, $111 = 3 \times 37$. Therefore $S$ is always divisible by $3$ and by $37$.Also, $S$ is exactly $111$ times $(X+Y+Z)$, so it is always divisible by $(X+Y+Z)$.Thus statement 1 (divisible by 3 and $(X+Y+Z)$) is correct, and statement 3 (divisible by 37) is correct.Statement 2 claims $S$ is always divisible by 9. Since $111$ is not a multiple of $9$, and $(X+Y+Z)$ need not supply the missing factor of $3$, $S$ is not always divisible by $9$. For example, take $X=1$, $Y=2$, $Z=4$: $S=111(7)=777$, which is not divisible by $9$. Hence statement 2 is false.So only statements 1 and 3 are correct.Answer:Option D (1 and 3).