jmc

We're really looking for the remainder when $8^{25}+12^{25}$ is divided by 100. Notice that $8=10-2$ and $12=10+2$. Then notice that the $k^{th}$ term of the expansion of $(10+2{)}^{25}$ is, by the binomial theorem, $\left(\genfrac{}{}{0ex}{}{25}{k}\right)\cdot 10^{25-k}\cdot 2^k$. Similarly, the $k^{th}$ term of the expansion of $(10-2{)}^{25}$ is, by the binomial theorem, $\left(\genfrac{}{}{0ex}{}{25}{k}\right)\cdot 10^{25-k}\cdot (-2{)}^k=(-1{)}^k\cdot \left(\genfrac{}{}{0ex}{}{25}{k}\right)\cdot 10^{25-k}\cdot 2^k$, which is the same as the $k^{th}$ term of $(10+2{)}^{25}$ for $k$ even, and the negative of the $k^{th}$ term of $(10+2{)}^{25}$ for $k$ odd. So, if we add together the $k^{th}$ terms of the expansions of $(10-2{)}^{25}$ and $(10+2{)}^{25}$, we get double the value of the $k^{th}$ term of the expansion of $(10+2{)}^{25}$, that is, $2\cdot \left(\genfrac{}{}{0ex}{}{25}{k}\right)\cdot 10^{25-k}\cdot 2^k$, if $k$ is even, and 0 if $k$ is odd. So, $8^{25}+12^{25}$ is the sum of all terms of the form $2\cdot \left(\genfrac{}{}{0ex}{}{25}{k}\right)\cdot 10^{25-k}\cdot 2^k$ for $0\le k\le 25$, $k$ even. But notice that this is divisible by 100 for $k<24$, and because we care only about the remainder when dividing by 100, we can ignore such terms. This means we care only about the term where $k=24$. This term is $$2\cdot \left(\genfrac{}{}{0ex}{}{25}{24}\right)\cdot 10^1\cdot 2^{24}=2\cdot 25\cdot 10\cdot 2^{24}=500\cdot 2^{24},$$which is also divisible by 100. So, $8^{25}+12^{25}$ is divisible by 100. So the sum of the last two digits is $0+0=\boxed{0}.$