The powers $$2^n$$ and $$5^n$$ start with the same digit where $$n$$ is a positive integer. What are the possible values for this digit?

[SOLVED]

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Rino Raj solved this puzzle:

Let the leading digit be $$x \in \{1, 2, \dots, 9\}$$. We have

$10^a x \lt 2^n \lt 10^a (x + 1) \\ 10^b x \lt 5^n \lt 10^b (x + 1)$

for some non-negative integers $$a$$ and $$b$$. Multiplying these inequalities and dividing by $$10^{a + b}$$, we have

$1 \le x^2 \lt 10^{n - a - b} \lt (x + 1)^2 \le 100.$

Since $$1 \lt 10^{n - a - b} \lt 100$$, $$n - a - b = 1$$. Thus

$1 \le x^2 \lt 10 \lt (x+1)^2 \le 100.$

The only digit $$x$$ for which this inequality holds true is $$x = 3$$. For example, $$2^5 = 32$$ and $$5^5 = 3125$$.

Credit

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