UPSC CDS 1 2020 Math Paper

Concept:A three-digit number $abc$ and its reverse $cba$ both divisible by 7 implies $99(a-c)$ is a multiple of 7.Since 99 is not divisible by 7, $a-c$ must be divisible by 7.Given all digits are different, $a-c$ cannot be 0, so $a-c=7$.Explanation:Let the number be $100a+10b+c$ and its reverse $100c+10b+a$.Both are multiples of 7, so subtract: $(100a+10b+c)-(100c+10b+a)=99(a-c)$ must be divisible by 7.Since 99 is not a multiple of 7, $a-c$ must be a multiple of 7.As $a$ and $c$ are first and last digits of a 3-digit number, $a$ can be 1–9 and $c$ can be 0–9, but $c\neq0$ for a 3-digit number? Actually $c$ can be 0, but then reverse would be a 2-digit number? Wait, reverse $cba$ must also be a 3-digit number, so $c\neq0$. So $a$ and $c$ range 1–9 with $a>c$.The only possible difference divisible by 7 is $a-c=7$ (since $a-c=0$ would give $a=c$, not allowed).Hence $a=9,c=2$ or $a=8,c=1$.Now test all digits $b$ from 0–9 such that $abc$ is divisible by 7 and all three digits are distinct.For $a=9,c=2$: numbers $9b2$. Only $b=5$ works (giving 952, divisible by 7). Reverse is 259.For $a=8,c=1$: numbers $8b1$. Only $b=6$ works (giving 861, divisible by 7). Reverse is 168.Thus the four numbers are 168, 259, 861, 952.Answer:Four (Option C).