Hey @eeveelution ! Sorry for replying so late, but here's how I would look at your two problems:

For the second one, since the problem only cares about the last digit of that huge sum, we only need to look at the last digit of each individual term!

\(1^{15}\) is easy, the last digit of \(1^{15}\) is just \(1\).

For the rest though, we don't want to expand it out (that's a lot of work).

So, for \(2^{15}\), let's rewrite it as \((2^5)^3\).

\(2^5=32 \implies \text{last digit is 2}\)

\(2^3=8\)

So the last digit of \(2^{15}=(2^5)^3\) is \(8\).

Another way to look at it is to try and find a pattern among powers of 2. Let's do a few and see if we can find anything.

\(2^1 \implies \text{last digit 2}\)

\(2^2 \implies \text{last digit 4}\)

\(2^3 \implies \text{last digit 8}\)

\(2^4 \implies \text{last digit 6}\)

\(2^5 \implies \text{last digit 2}\)

\(2^6 \implies \text{last digit 4}\)

Wait a minute, this is just gonna repeat \(2,4,8,6,2,4,8,6...\) It cycles every \(4\)!

So, since \(15\) is \(3\) more than a multiple of \(4\), the last digit of \(2^{15}\) is going to be the \(3^{\text{rd}}\) number in the list, which is \(8\).

You can repeat this process for the rest of the numbers.

But, keep in mind that \(12\) has units digit \(2\), so \(12^{15}\) is going to have the same units digit as \(2^{15}\)!

Also, one last tip-- since \(2012\) is so big, I would group everything in groups of \(10\). What I mean is, first find what the units digit of \(1^{15}+2^{15}+...+10^{15}\) is. Then, since \(11^{15}\) through \(20^{15}\) will have the same units digits (along with every other group of 10), you can use a shortcut and not have to calculate every single one.

For the rectangle problem, I'm getting 79... which is weird... but either way I definitely think it's more than 73. What book is this?