It's just simple number theory
A positive integer is written on a chalkboard. We repeatedly erase its unit digit and add five times that digit to what remains. If this integer is , can we ever end up at ?
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1 solution
No. To see this, let = denote congruence. Then each step can be generalized as 10x + y -> x + 5y. Now suppose 10x + y = 0 (mod 7). y = -10x = 4x (mod 7) It follows that x + 5y = x + 5(4x) = 21x = 0 (mod 7). That is, if we start with a number which is divisible by 7, then we'll always have a number divisible by 7. Obviously 7^2007 is divisible by 7, but 2007^7 = (-2)^7 = -128 = 5 (mod 7) so it isn't divisible by 7.