Millionth powers

The Endeavour 2025-04-17

I was poking around Richard Stanley’s site today and found the following problem on his miscellaneous page.

Find a positive integer n < 10,000,000 such that the first four digits (in the decimal expansion) of n^1,000,000 are all different. The problem should be solved in your head.

The solution is not unique, but the solution Stanley gives is n = 1,000,001. Why should that work?

Let M = 1,000,000. We will show that the first four digits of (M + 1)^M are 2718.

$\begin{align*} (1 + M)^M &= \left(M\left(1 + \frac{1}{M}\right)\right)^M \\ &= M^M \left(1 + \frac{1}{M}\right)^M \\ &\approx M^M e \\ &= 2718\ldots \end{align*}$

This uses the fact that (1 + 1/n)ⁿ → e as n → ∞. If you’re doing this in your head, as Stanley suggests, you’re going to have to take it on faith that setting n = M will give you at least 4 decimals of e, which it does.

If you allow yourself to use a computer, you can use the bounds

$\left(1 + \frac{1}{n}\right)^n < e < \left(1 + \frac{1}{n}\right)^{n+1}$

to prove that sticking in n = M gives you a value between 2.718280 and 2.718283. So in fact we get 6 correct decimals, and we only needed 4.

There are many solutions to Stanley’s puzzle, the smallest being 4. The first four digits of 4^M are 9802. How could you determine this?

You may not be able to compute 4^M and look at its first digits, depending on your environment, but you can tell the first few digits of a number from its approximate logarithm.

log₁₀ 4^M = M log₁₀ 4 = 602059.9913279624.

It follows that

4^M = 10⁶⁰²⁰⁵⁹ 10^0.9913279624 = 9.80229937666385 × 10⁶⁰²⁰⁵⁹.

There are many other solutions: 7, 8, 12, 14, 16, …

The post Millionth powers first appeared on John D. Cook.

Millionth powers

The Endeavour 2025-04-17

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