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 \(5^5 = 3125\).

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