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Can anyone help me find the first digit of $2015^{2015}$?

It is easy to find the last digit but I have no idea for the first digit.

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  • 5
    $\begingroup$ Use logarithms to approximately calculate the result. $\endgroup$ – Mario Carneiro Feb 6 '15 at 14:19
  • $\begingroup$ Is this a question from an on-going contest? $\endgroup$ – Joel Reyes Noche Mar 17 '15 at 4:20
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Now, you didn't say "by hand" as these calculations often do, but I'll assume that at least you're from before 1950 so that solutions like just asking Wolfram Alpha about a mere 8000-digit number are beyond you.

We can calculate $\log_{10}(2015^{2015})=2015(\log_{10}(2.015)+3)$, and consulting my trusty tables I find that $$\log_{10}(2.015)\approx\frac{\log_{10}(2.01)+\log_{10}(2.02)}2\approx\frac{0.30320+0.30535}2\approx0.30428,$$ so $\log_{10}(2015^{2015})\approx6658.1$. (I need to do this calculation well enough to show it is between $6658$ and $6658+\log_{10} 2\approx6658.3$ to prove the claim.)

Then, $6658\le\log_{10}(2015^{2015})<6658+\log_{10} 2$ gives $10^{6658}\le2015^{2015}<2\cdot 10^{6658}$, so the first digit is $1$.

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    $\begingroup$ $1$ is the most likely leading digit anyway :) $\endgroup$ – Hagen von Eitzen Sep 23 '15 at 18:29
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The units digit is of course a 5.

For the leading digit, see that $n^n = 10^{n\cdot log_{10}(n)} \approx 10^{2015\cdot3.304275} \approx 10^{6658.114} = 10^{6658}10^{0.114} \approx 1.3008\cdot 10^{6658}$. Therefore, the first digits of your number are 13008...

Check: http://www.wolframalpha.com/input/?i=2015^2015

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    $\begingroup$ I wasn't sure which digit the OP was referring to, so I wrote about both. Usually to avoid confusion I call them "leading" and "units" digit. $\endgroup$ – Luigi D. Feb 6 '15 at 14:45
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Take the base-ten logarithm.
By the log laws, that will equal $2015\log2015$. Suppose that is $x+y$, where $x$ is a whole number and $0<y<1$. The number has $x+1$ digits.
$y$ tells you the first digit. Or rather, $10^y$ starts with the same digits $2015^{2015}$ does.

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