1998 RMO

A positive integer is written on a board. We repeteadly erase its unit digit and add 5 5 times that digit to what remains. Starting with 7 1998 7^{1998} , can we ever end up at 199 8 7 1998^7 ?

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1 solution

Mark Hennings
May 29, 2019

Suppose that a number n n has final digit b b . Then n = 10 a + b n = 10a+b for some integer a a . After rubbing out the unit digit b b and adding 5 b 5b to what remains, we have n ^ = a + 5 b \hat{n} = a+5b . Since n ^ + 2 n = ( a + 5 b ) + 2 ( 10 a + b ) = 7 ( 3 a + b ) \hat{n} + 2n = (a+5b) + 2(10a + b) = 7(3a + b) , if n n is a multiple of 7 7 , then so is n ^ \hat{n} . If we start with 7 1998 7^{1998} , any subsequent number obtained by this process must be a multiple of 7 7 , and 199 8 7 1998^7 is not a multiple of 7 7 . Thus we can never end up with 199 8 7 1998^7 .

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