We raise the numbers 2 and 5 to the same positive integer power and get two numbers that have the same first (leftmost) digit. What are the possible values of the first digit?
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Well... I believe the only possible number is 3. – Samuel Mar 28 '17 at 16:35
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First as in leftmost or rightmost? – John Mar 28 '17 at 16:37
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@John: for the rightmost the only answer is $1$ with the power being $0$. – Ross Millikan Mar 28 '17 at 16:42
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I don't know how to find the first digit... If it were possible to have the total number n of digits (length of number) and then divide by 10^(n-1) and then round down the quotient... – Samuel Mar 28 '17 at 16:44
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Just thinking out loud: We can also take the logarithm "L" of the number, take its integer part "i" and then divide by 10^i, then round down the quotient. I don't know if any of these things make sense! – Samuel Mar 28 '17 at 16:53
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@RossMillikan 10^n. So we express 5 as 10/2? And then raise to the same power? – Samuel Mar 28 '17 at 17:55
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That is correct. Now you need two numbers with the same first digit that multiply to $10^n$. One choice is $100 \times 1000$ or something similar, but those are never powers of $2$ and $5$. Can you find another choice? If you don't see it right away, just start listing $2^n, 5^n$ and see when they start with the same digit. – Ross Millikan Mar 28 '17 at 17:57
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@RossMillikan I see they have the same first digit when the power is 5, 15, 78, 88, 98, 108 and so on. – Samuel Mar 28 '17 at 18:04
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...but I still don't get it :( :( – Samuel Mar 28 '17 at 18:30
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We know that the leading digits of $2^n$ and $5^n$ are the same and that the product starts with $1000\ldots$. If they leading digits of $2^n$ and $5^n$ were $6$, for example, we would have $36\ldots \le 2^n5^n \lt 49 \ldots$. The only leading digit that includes $1000\ldots$ is $3$ because we know the product is between $9\ldots $ and $16\ldots$ We need the leading digit to be about $\sqrt {10}$ and $3$ qualifies. – Ross Millikan Mar 28 '17 at 21:27
