47th Putnam 1986 Problem A2 What is the remainder when the integral part of 10 20000 /(10 100 + 3) is divided by 10? Solution
: 3. Easy. We start with the identity 1 = ( 1 - (3/10 100 ) + (3/10 100 ) 2 - (3/10 100 ) 3 + ... ) (1 + 3/10 100 ). Multiplying through by 10 20000 and dividing by (10 100 + 3) we get: 10 20000 /(10 100 + 3) = ∑ {10 19900 (-3/10 100 ) n } (*), and the early terms in this sum are all integral. In fact they are integral for n < 200 and these terms sum to 10 19900 (1 - (3/10 100 ) 200 ) /(1 + 3/10 100 ) = (10 20000 - 3 200 )/(10 100 + 3). Multiplying by (10 100 + 3) and subracting from 10 20000 gives 3 200 = 9 100 which is less than 10 100 + 3. So the integral part of 10 20000 /(10 100 + 3) is indeed given by the first 200 terms, 0 ≤ n < 200, of (*). The first 199 terms are all divisible by 10. The last term is (-3) 199 . But 3 2 = -1 (mod 10), so (-3) 198 = 3 198 = (-1) 99 = -1 (mod 10), and hence (-3) 199 = 3 (mod 10). 47th Putnam 1986 © John Scholes jscholes@kalva.demon.co.uk 2 Oct 1999