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In the book$^1$ that I am reading, the author dubs an $n$-digit positive number a Smallbrain number if it is equal to the sum of the $n$th powers of its digits, with $371 = 3^3 + 7^3 + 1^3$ given as an example. The question is to prove that there are no Smallbrain numbers with 1000 digits.

The problem is that for $n = 1000$, the sum of the $n$th powers of the digits is at most $1000\cdot9^{1000}$, which is strictly less than the least number with 1000 digits, namely $10^{999}$. I showed this as follows:

$$(*) \iff 9^{1000} < 10^{996}$$

$$\impliedby 9^{1000} < 10^{990}$$

$$\impliedby 9^{10} < \bigg(1+\frac{1}{9}\bigg)^{990}$$

$$\impliedby 9^{10} < 2^{110}$$

using that $\bigg(1+\frac{1}{n}\bigg)^{n} \ge 2.$

This question, however, is printed immediately the after a chapter explaining the AM-GM inequality and Cauchy's inequality for $n = 2$. Although I realise that the fact I used can be derived from AM-GM, I doubt that this was expected for the answer. Is there are a more direct way to prove $(*)$ using the two inequalities I mention?

1: A Concise Introduction to Pure Mathematics, Liebeck.

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  • $\begingroup$ Well, $1000\ln(9)= 2197.2245<2293.37<996\ln(10)$ is pretty easy, unless there is some reason this needs to be done with pencil and paper. $\endgroup$
    – lulu
    Commented Jul 29, 2021 at 18:11
  • $\begingroup$ @lulu Sure, but my method is already pretty easy and it only used pencil and paper. I'm just wondering if there is a way to do it using AM-GM/Cauchy since those were the topics discussed in the previous chapter for which this question is an exercise. Logarithms have not been mentioned in the book yet. $\endgroup$
    – legionwhale
    Commented Jul 29, 2021 at 18:18
  • $\begingroup$ Alternatively; from $5^3<2^7$ it follows that $5^{10}<10^7$ and so $250^5<10^{12}$. Then from $3^5<250$ it follows that $3^{25}<10^{12}$ and so $9^{1000}<10^{960}$. $\endgroup$
    – Servaes
    Commented Jun 29 at 17:50

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Weighted AM-GM can solve this pretty easily: $$9^{1000} < 10^{996} \iff 9^{250}<10^{249} \iff 9 <\left(\frac{10}{9}\right)^{249} \iff 9^{\frac{1}{249}} <\frac{10}{9}$$ Note that by AM-GM on $9,1$ with weights $1, 248$ respectively, we get $$9^{\frac{1}{249}}<\frac{9+248}{249}<\frac{10}{9}$$ The last inequality can be easily verified by cross multiplication.

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