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Hello fellow mathematicians,

I have encountered a problem that I could not solve in Justin Stevens's book on Olympiad number theory. The question is as follows:

"Let $v(n)$ be the number of factors of $2$ in the number $n!$. What is the least value of $n$ such that $n-v(n)=1990?$"

Any help on this question would be extremely helpful as this is puzzling me greatly.

Thank you in advance.

I have been experimenting with small values of $n$ and have noticed a slow increase in the value of $n-v(n)$, and this value seems to be 1 whenever $n$ is a power of 2, and the pattern 'resets', as the value of $n-v(n)$ becomes small again.

IChoi
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    How far have you calculated $v(n)$ and $n-v(n)$? When a random largish number appears in a contest problem, one of the first things you should do is to check out small cases, wishing to get enlightenment. The fact that you ask here rather than try and do anything yourself is a cause for concern. If you want to do well in a math contest you have to be prepared to see problems where you don't know how to proceed. That's the point of having those contests. – Jyrki Lahtonen Apr 15 '22 at 04:15
  • Mind you, the formula for $v(n)$ has appeared on this site frequently enough. Hint: search for questions asking how many zeros are at the end $n!$ for some largish $n$. – Jyrki Lahtonen Apr 15 '22 at 04:17
  • Another hint: what do you notice about the numbers $n$ such that $n-v(n)=1$? What about those with $n-v(n)=2$? Of course, having the formula for $v(n)$, suitably reinterpreted, makes the question trivial, but... – Jyrki Lahtonen Apr 15 '22 at 04:21
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    I know the Legendre formula for calculating v(n) for factorials - I just don't know how to solve this particular question. I've tried the cases for small values of n, up to 10, and there seems to be slow increase of the value of $n-v(n)$, and it seems to be related to powers of 2, but I could not find an obvious pattern. I've spent about an hour pondering and experimenting before finally asking for some guidance on the site. – IChoi Apr 15 '22 at 05:03
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    IChoi, good. Powers of two, indeed, play a key role. That is the kind of observation I was looking for. Keep going, at least up to 16, I think! And please add a summary of your findings to the question body. – Jyrki Lahtonen Apr 15 '22 at 05:08
  • Thanks for the edit. Retracting my vote to close. – Jyrki Lahtonen Apr 15 '22 at 05:18
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    Thanks for your help Jyrki Lahtonen - I have continued experimenting until 16 and it seems like, while the numbers in between the powers of 2 don't have a clear pattern, $2^{n}-1 - v(2^{n}-1)=n$. I think I'll try proving this conjecture. – IChoi Apr 15 '22 at 05:20
  • Great! You are half way there :-) – Jyrki Lahtonen Apr 15 '22 at 05:21
  • When you are done, you can take a peek at Wikipedia. Locally for example here or here. – Jyrki Lahtonen Apr 15 '22 at 05:35
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    Thank you very much - I've proved it and have reached an answer. I should have gone a bit further before giving up on experimenting. Thanks for your help again. – IChoi Apr 15 '22 at 05:47
  • Great! It is totally ok to also post the solution as an answer. That way you may get more feedback on the details (if anyone is so inclined). Your call, of course. – Jyrki Lahtonen Apr 15 '22 at 05:59

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I will change the notation to standard notation such that $\nu_2(n!)$ denotes the largest power of $2$ that divides $n!$. The key here is the result that $\nu_2(n!)=n-s_2(n)$, where $s_2(n)$ denotes the sum of digits in base $2$. From there it isn't hard to finish.

Joel Gerlach
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