last three digits of $7^{100}-3^{100}$ How can I find the las three digits of $7^{100}-3^{100}$ ? I know one way is to use $7^{100}=(10-3)^{100}=\sum_{k=0}^n{100 \choose k}10^{100-k}(-3)^k$ but I'm totally stuck... 
Thanks

Well, note that for $0\le k\le 97,$ we have that $1000$ is readily a factor of $\binom{100}{k}10^{100-k}(-3)^k,$ so the last three digits of all of those numbers will be $0$s. Hence, the last three digits of $7^{100}$ will be the last three digits of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}10^1(-3)^{99}+\binom{100}{100}10^{0}(-3)^{100},$$ that is, of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}(-3)^{99}+3^{100}.$$ Hence, the last three digits of $7^{100}-3^{100}$ will be the last three digits of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}10^1(-3)^{99}.$$
Since $\binom{100}{98}=50\cdot99$ and $\binom{100}{99}=100$, the last three digits of this number are easily found.
