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In a ΜAΘ contest from 1991, I found this problem in my problem book. I know how to solve problems like this, and I know how to solve it if the problem tells me to find the digits in $2^{44}$, but $5^{44}$ makes me think about the problem like this:

$${5^{44} = \frac{10^{44}}{2^{44}}}$$

Given this, I don't think I can begin to approximate it with a change of base, since I'm only given ${\log_{10} 2}\approx {0.3010}$, not ${\log_2 10}$ which I could calculate. Maybe I could try $44$ digits from the $10^{44}$ minus the digits in $2^{44}$?

$${44} - 2^{44\log_210}$$

I don't think thats how division works either, I'm very lost. Please do help.

Gary
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Coffeelicious
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  • It's one of those tricks one picks up and then pretends one figured out on ones one: A number $n$ has $k$ digits if and only if $10^{k-1}\le n < 10^{k}$ if and only if $k-1 \le \log_{10} n < k$. – fleablood Mar 28 '24 at 06:31

1 Answers1

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You are on the right track.

Note that we want to compute $$\log_{10}(5^{44}) = 44\log_{10}(5) = 44 \log_{10}(10 / 2) = 44(\log_{10}(10) - \log_{10}(2)) = 44(1 - 0.3010) \approx 30.7$$ so we get 31 digits.

daisies
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  • How do you determine whether it's 30 or 31 from 30.7? – Gwen Mar 28 '24 at 05:49
  • The logarithm is increasing, so if $\log_{10}(10^{30}) = 30$ and $\log_{10}(10^{31}) = 31$, then a number with logarithm 30.7 has 31 digits (the same amount of digits as $10^{30}$). I made a typo in the answer. Sorry! – daisies Mar 28 '24 at 05:57