If the leading digit of a positive integer is deleted, the number gets reduced by 57 times. Find all such numbers.
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$$57\cdot n=10^r m+ n\iff56n=10^r m$$
$\iff8n=\dfrac{10^rm}7\implies 7\mid m, m=7q$(say)
As $0<m<10, m=7$
$\implies8n=10^r\implies r\ge3, n=5^r2^{r-3}$
lab bhattacharjee
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@Hanzala, Put a few values of $r\ge3$ you should understand the steps – lab bhattacharjee Aug 20 '16 at 10:59