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I have been trying to work through the following exercise:

Find the power of $5$ in the prime factorization of $2020!$.

So far I have worked out that the prime factorization of $2020$ is $2^2 \cdot 5^1 \cdot 101^1$, however I am not sure if this is useful or not!

It may also be useful to note that this is an example from a number theory course and so I believe there should be a methodical process to find the answer, however I have been unsuccessful so far!

I have pondered whether maybe I could use the fact that $5^4$ is the highest power of $5$ that is less than $2020$.

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    $\begingroup$ Find the power of $5$ in, say, the first $25$ factorials. It'll go faster than you think and possibly tell you all you need to know, if you try to be systematic about it. $\endgroup$
    – Arthur
    Commented Mar 29, 2021 at 21:22
  • $\begingroup$ @Arthur I'll try that now! $\endgroup$
    – mathemagic
    Commented Mar 29, 2021 at 21:25
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    $\begingroup$ @mathemagic Cf. Legendre's formula. $\endgroup$
    – John Omielan
    Commented Mar 29, 2021 at 21:37
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    $\begingroup$ With respect to the article cited in John Omielan's comment, pay particular attention to the proof(s). It is well worth experminenting with small numbers [e.g. $v_5(100!) = 24$] to understand why the formula is accurate. $\endgroup$
    – user2661923
    Commented Mar 29, 2021 at 22:41

1 Answer 1

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Let $k$ be the exponent of $5$ in the prime factorization of $2020! = 1\cdot 2\cdot \dots\cdot 2020$. Each factor in the set $A_1=\{5, 10, 15, \dots, 2015, 2020\}$ contributes $1$ to $k$. On top of that, each factor in the set $A_2=\{25, 50, \dots, 1975, 2000\}$ contributes $1$ to $k$. On top of that, each factor in the set $A_3=\{125, 250, \dots, 1875, 2000\}$ contributes $1$ to $k$. Continuing this reasoning we see that

$$ k = \sum_{i=1}^\infty|A_i| = \sum_{i=1}^\infty \left\lfloor\frac{2020}{5^i}\right\rfloor = 404 + 80 + 16 + 3 + 0 + 0 + \dots= 503. $$

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