Pranav picks a three-digit number and subtract it from the same number after reversing the order of its digits. If this difference is divided by 9. Find the number.

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Solution

let number be $a b c$ where $a < c$
on reversing the order the number becomes $c b a ⟹ c b a > a b c$
on subtracting $c b a - a b c$ we get $10 + a - c$ in ones place $9$ in tens place and $c - a - 1$ in hundreds place
for a number to be divisible by $9$ the sum of the digits should be divisible by $9$
$⟹ 10 + a - c + 9 + c - a - 1 = 18$ is always divisible by $9$
$∴$ every three digit numdber satisfies the given condition

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