What is the remainder when the integral part of 10^{20000}/(10^{100} + 3) is divided by 10?

**Solution**

Answer: 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).

© John Scholes

jscholes@kalva.demon.co.uk

2 Oct 1999