The Mathematical Ninja placed his quadruple espresso on the table; @dragon_dodo looked up from her laptop and smiled. “You’re taking it easy on the caffeine this morning, I see?”

The Mathematical Ninja nodded. “Yeah, this is only my third cup. What are you working on?”

“Computing assignment.”

The Mathematical Ninja, luckily, had already drained his coffee and couldn’t immediately spit it out in disgust.

“Find the largest seven-digit Fibonacci number.”

“$F_{35}$ is about 9.2 million,” said the Mathematical Ninja.

“Have you memorised the first three dozen Fibonacci numbers, or did you work that out?”

He sighed, gestured at the barista for another coffee. “I worked it out, of course. I’ve got better things to do with my memory.”

“Go on.”

“Like, for example, remember Binet’s formula, which is $F_{n} = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}$.”

“I suppose that’s handy if you know your powers of $\phi$ forwards and backwards, and how to divide quickly by $\sqrt{5}$. Which I imagine you do.”

“One out of three ain’t bad,” said the Mathematical Ninja. “Although, tut-tut: knowing them backwards is hardly necessary: $(-\phi)^{-35}$ is so small as to make no difference. The one I know,” he anticipated, “is dividing by $\sqrt{5}$, but we’ll get to that.”

“So you do something clever to work out $\phi^{35}$, presumably?”

“But of course. First, though, we need to know why it’s 35. We need $\frac{\phi^{n}}{\sqrt{5}} \lt 10^7$, which is the smallest 8-digit number.”

“Take logs, of course. $n \ln(\phi) \lt 7 \ln(10) + \frac 12 \ln(5).$”

“Correct.” He sipped the coffee this time; he was getting to the point where he could see through it. “$\phi$ is a shade more than $\frac{8}{5}$, so $\ln\left(\phi\right) \approx 2.1 - 1.6 = 0.5$. In fact it’s a bit smaller, and I knocked it down to 0.48.”

“It’s 0.481212,” said @dragon_dodo. “I had it up on the screen already.” She thought she was probably safe from any repercussions, but bought him another coffee just to be on the safe side. “The right hand side is about $7 \times 2.3 + 0.8 = 16.9$.”

“Indeed, so we need to find the largest multiple of $\ln(\phi)$ that goes into that.”

“34 almost certainly does – but $35 \times 0.48 = 70 \times 0.24 = 16.8$, so that’s even better!” @dragon_dodo grinned. “So you worked out $e^{16.8}$?”

“Almost,” said the Mathematical Ninja – “remember that that’s $\phi^{35}$, and we’ve still to divide by $\sqrt{5}$.”

“Aha! So it’s actually about $e^{16}$?”

“Yeeees… although I’d be a bit more careful. Small differences in the power make big differences in the result, so I’d do $35 \times 0.4812 - 0.5 \times 1.609$, which is $70 \times 0.2406 - 0.805$, or 16.037.”

“More than a third of a per cent!” said @dragon_dodo. “But let’s see, I know that $e^3$ is a bit more than 20, so $e^{15}$ is about 3 million, so somewhere about 9 million looks good.”

“I actually switched to base 10,” said the Mathematical Ninja, looking slightly ashamed. “I know that 2.303 × 7 must be in the same ballpark – it’s 16.121 – which means I need to lose 7 or 8%. When in doubt, round even: 9.2 million.”

@dragon_dodo tapped at the computer for a moment.

“Don’t you dare claim you’re just checking Twitter,” said the Mathematical Ninja, smugly reading 9,227,465 through the screen.

* Edited 2015-02-22 to fix a typo.