# A Challenge to the Mathematical Ninja

“I *beg* your pardon?!” yelled the Mathematical Ninja.

The terribly well-dressed gentleman stood his ground. “I said, sensei, I would work $0.8^{10}$ out differently.”

A sarcastic laugh. “This, I have to see!”

“Well, $8^{10} = 2^{30}$, which is about $10^{9}$.”

“*About*.”

“Obviously, we can do better with the binomial: $2^{10}$ is 1024, so we can split up $2^{30}$ as $\left(1000 + 24\right)^3$. That’s $10^{9} + 3\times 24 \times 10^6 + 3\times 24^2 \times 10^3 + 24^3$.”

“That’s easy for you to say.”

“1,072 million is easy enough. $3 \times 24^2$ is $72 \times (25-1)$, or $1800 - 72 = 1728$, giving us 1,073,728 thousands.”

“I would have noted that $3 \times 24^2 = 12^3$.”

“And… I mean, we’re already good to 5sf, but we could get it exactly: we can approach $24^3$ as $24 \times (25-1) \times (25-1)$ - which is $(600 - 24) \times (25-1)$. Six hundred 25s are 15,000, less 1200 plus 24, making 13824. Adding on, that’s 1,073,741,824 - and all we need to do is put the decimal point in the right place.” A small nod, a small straightening of the bow-tie.

The Mathematical Ninja’s eyebrows furrowed. “Yes.” Ahem. “Yes, I was wondering who’d be the first to spot that.”

* Many thanks to Adam Atkinson and CyberChalky, who independently suggested the same thing.

* Edited on 2019-06-22 to correct errors, and on 2019-06-23 to amend the correction.

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