What is the remainder when the integral part of 1020000/(10100 + 3) is divided by 10?
Solution
Answer: 3.
Easy.
We start with the identity 1 = ( 1 - (3/10100) + (3/10100)2 - (3/10100)3 + ... ) (1 + 3/10100 ). Multiplying through by 1020000 and dividing by (10100 + 3) we get: 1020000/(10100 + 3) = ∑ {1019900 (-3/10100)n} (*), and the early terms in this sum are all integral. In fact they are integral for n < 200 and these terms sum to 1019900(1 - (3/10100)200) /(1 + 3/10100) = (1020000 - 3200)/(10100 + 3). Multiplying by (10100 + 3) and subracting from 1020000 gives 3200 = 9100 which is less than 10100 + 3. So the integral part of 1020000/(10100 + 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 32 = -1 (mod 10), so (-3)198 = 3198 = (-1)99 = -1 (mod 10), and hence (-3)199 = 3 (mod 10).
© John Scholes
jscholes@kalva.demon.co.uk
2 Oct 1999