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Answer: $2^{13}$
$$\begin{align} &\left(64 \text{ men }+64\text{ women}\right)\times 64\text{p}\\
2^3 = 8,\hspace{2mm} 2^4 = 16,\hspace{2mm} &2^5 = 32,\hspace{2mm} 2^6 = 64\\
&\left(\hspace{6mm}2^6 \hspace{8mm}+\hspace{8mm} 2^6 \hspace{6mm}\right)\times 2^6\\
=&\left(2^6\times2\right)\times2^6\\
=&2^6\times2^1\times2^6\\
=&2^{13}\end{align}$$
Maundy Money
Age 11 to 14
ShortChallenge Level 
Answer: $2^{13}$
$$\begin{align} &\left(64 \text{ men }+64\text{ women}\right)\times 64\text{p}\\
2^3 = 8,\hspace{2mm} 2^4 = 16,\hspace{2mm} &2^5 = 32,\hspace{2mm} 2^6 = 64\\
&\left(\hspace{6mm}2^6 \hspace{8mm}+\hspace{8mm} 2^6 \hspace{6mm}\right)\times 2^6\\
=&\left(2^6\times2\right)\times2^6\\
=&2^6\times2^1\times2^6\\
=&2^{13}\end{align}$$
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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