“Sensei! I have a problem!”

The Mathematical Ninja nodded. “Bring it on.”

“There’s a challenge! Someone has picked a five-digit integer and cubed it to get 6,996,364,932,376. I know it ends with a six, and I could probably get the penultimate digit with a bit of work… I just wondered if there was a better way. Without a calculator, of course.”

The Mathematical Ninja hmmed for a moment. “Tricky.”

The student’s eyes boggled. This was not a word he had heard escape the Mathematical Ninja’s lips before.

“It’s a tiny bit less than $7 \times 10^{12}$,” said the Mathematical Ninja, thoughtfully. “Therefore the cube root is somewhere slightly below 20,000.”

This sounded plausible. “So the first digit is 1, the second digit is high, and the final digit is 6. We’ve halved the search space!”

“We’ve square rooted the search space,” corrected the Mathematical Ninja.

“How about we take logs, sensei?”

“Naturally.”

The student continued. “$\ln(7)$ is roughly 1.95, and - looking at the billions - we’re off of 7,000 by about 3.5, which is one part in 2000. And we don’t know $\ln(7)$ accurately enough for that to be worth adjusting for. $12\ln(10)$, though, is $12 \times 2.302$, which is 23.02 + 4.605, let’s say 27.63. Overall, we’re now at 29.58 - and we need to divide that by 3. A shade less than 10.”

“Indeed. 9.86, to be moderately precise.”

“OK. And we want to unlog that. So what’s $e^{9.86}$?”

“Let’s break it down. We already know it’s a bit more than $10^4$, the natural log of which is…”

“$4 \times 2.302 = 9.208$,” said the student. “Call it 9.21?”

“That’s for the best,” said the Mathematical Ninja. “After all, it’s 9.2103.”

The student’s eyes narrowed slightly. For all his respect for the Mathematical Ninja, he of all people didn’t like a smartarse. “So we’ve got 0.65 left to account for - a bit less than 2, as you said. But how much less? 0.04, if I remember correctly.”

“You do,” said the Mathematical Ninja.

“So, 4% less than 2 is 1.92, meaning our answer is probably 191_6 or 192_6.”

“It would be hard to disagree,” agreed the Mathematical Ninja.

“Should I just try cubing all of the numbers between 19,106 and 19,296 that end in 6?”

“That would doubtless work,” said the Mathematical Ninja, but with a sneer that suggested to the student that there may be a better way.

“I suppose there may be a better way,” said the student.

“You might start by dividing 6,996,364,932,376 by 8, reasoning that it’s an even cube. You get 874,545,616,547.”

“Is that much better?”

“It tells you the last digit is 3, which means the penultimate digit of what you had must be even. Also, the digital root of the original cube is 1, which means the digital root of the cube root must be 1, 4 or 7.”

“That narrows it down significantly! So it could be 19,126, 19,186, 19,246… and that’s it!”

“Mhm,” said the Mathematical Ninja.

“Do I still need to cube those?”

“Sure. But be lazy. Binomial expansion. Modulo 100.”

“You mean something like $(10a+b)^3$?”

“Something very like that. Identical, in many respects.”

Another eye-narrowing moment passed. “So, $1000a^3 + 300a^2b + 30ab^2 + b^3$, which has to end in 76. The first two terms vanish modulo 100, and $b=6$, so I’ve got $30 \times 36 a + 216$ giving 76 (mod 100). Strip out the hundreds, that’s $80a + 16$ giving 76 (mod 100), and $80a = 60$. So $a$ has to be… 2!”

“Bingo,” beamed the Mathematical Ninja.

“19,126. Can I try that on the calculator to be sure?”

“Do it longhand,” said the Mathematical Ninja. “Don’t push your luck.”