$\begin{align}&2^6+2^5+2^4+2^4\\

=&2^6 + 2^5 + 2^4\times2\\

=&2^6 + 2^5 + 2^{4+1}\\

=&2^6 + 2^5 + 2^5\\

=&2^6 + 2^5\times2\\

=&2^6 + 2^6\\

=&2^7\end{align}$

$2^2=4$ $2^5=32$

$2^3=8$ $2^6=64$

$2^4=16$ $2^7=128$

So $2^6+2^5+2^4+2^4=64+32+16+16=128=2^7$

Notice that all of the numbers in the sum are multiples of $2^4$, since $2^6=2^2\times2^4,2^5=2\times2^4,2^4=1\times2^4.$ So

$$\begin{align}2^6+2^5+2^4+2^4&=2^2\times2^4+2\times2^4+1\times2^4+1\times2^4\\

&=\left(2^2+2+1+1\right)\times2^4\\

&=\left(4+2+1+1\right)\times2^4\\

&=8\times2^4\\

&=2^3\times2^4\\

&=2^7\end{align}$$

You can find more short problems, arranged by curriculum topic, in our short problems collection.