Question 430331
Suppose our n-digit integer is {{{x = 6*10^(n-1) + sum(a[i]10^i, i = 0, n-2)}}}. Then,


{{{sum(a[i]10^i, i = 0, n-2) = (6*10^(n-1) + sum(a[i]10^i, i = 0, n-2))/25}}} 


{{{25*sum(a[i]10^i, i = 0, n-2) = 6*10^(n-1) + sum(a[i]10^i, i = 0, n-2)}}} (multiply both sides by 25)


{{{24*sum(a[i]10^i, i = 0, n-2) = 6*10^(n-1)}}}


{{{sum(a[i]10^i, i = 0, n-2) = (10^(n-1))/4 = 25*10^(n-3)}}}


This simply looks for the (n-1)-digit numbers whose decimal representation is 25000...0. This means that the "remaining" number can be 25, 250, 2500, ... We put the 6 in front, so all integers 625, 6250, 62500, ... satisfy.