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What is the greatest power of 18 that divides 190! ?

Consider that $18 = 2\cdot3\cdot3$, hence it is a product of prime numbers. So we are seeking $\mu$ in $(2\cdot3\cdot3)^\mu$. We use De Polignac's formula:

$$\mu=\sum_{k>0}\lfloor\bigg(\frac{n}{p^k}\bigg)\rfloor$$

By the technique to solve it we consider that:

$$\lfloor\bigg(\frac{n}{p^{k+1}}\bigg)\rfloor=\lfloor\bigg(\frac{n}{p^k}\bigg)\cdot\frac1p\rfloor$$

We insert for the numbers, $n=190$, but $p$ is not prime, but a product of primes. Any ideas if there is a better alternative than this or if I can use this formula by re-writing the numbers?

Bill Dubuque
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Luthier415Hz
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    $2$ appears more often than $3$ in prime factorisations of consecutive integers, so we only have to count how many times $3$ divides $190!$. Then De Polignac's formula works, and we just divide by $2$ and floor it at the end (since $18$ is divisible by $3$ twice). – SomeCallMeTim Jul 18 '23 at 14:18
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    Yes, you are correct that 18 not being prime is a problem for De Polignac's formula... but 2 and 3 are both primes. Do each separately and then use that info to see how many 18s you can get from that. – DotCounter Jul 18 '23 at 14:18
  • @AmateurDotCounter do you mean solve $$\sum_{k>0}\lfloor\bigg(\frac{190}{3^k}\bigg)\cdot\frac13\frac12\rfloor$$ ? Using Mathematica, that converges only if 3^{-k}=0, which is a little bit difficult to understand. Which to me means that there does not exist any value of k that satisfies the question. – Luthier415Hz Jul 18 '23 at 14:30
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    That's a finite sum, since you have the floor. There is no need to check for $2$...clearly, $3$ is the limiting factor. – lulu Jul 18 '23 at 14:31
  • How do you do that, when the floor function is of a exponential number? – Luthier415Hz Jul 18 '23 at 14:33
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    @Luthier415Hz No, I mean $\mu_3=\sum_{k>0}\lfloor\bigg(\frac{190}{3^k}\bigg)\rfloor$ should tell you the exponent for 3... and a similar expression should get you the expression for the exponent of 2. – DotCounter Jul 18 '23 at 14:33
  • @AmateurDotCounter it appears by Mathematica that there is no value of k that satifies this. – Luthier415Hz Jul 18 '23 at 14:35
  • Not really, I have to solve this by Polignacs formulae – Luthier415Hz Jul 18 '23 at 14:40
  • Ah, I didn't see that explicitely. I see that now. This could be easier than I thought. I will provide a self.answer. – Luthier415Hz Jul 18 '23 at 14:47

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Starting from the DePolignac/Legendre equation we can solve, for prime $p=3$, what greatest exponent of $3^\mu$ properly divides $n=190!$. $$ \mu_3=\sum_{k>0}\bigg\lfloor\frac{190}{3^k}\bigg\rfloor=\bigg\lfloor\frac{190}{3^1}\bigg\rfloor+\bigg\lfloor\frac{190}{3^2}\bigg\rfloor+\bigg\lfloor\frac{190}{3^3}\bigg\rfloor+\bigg\lfloor\frac{190}{3^4}\bigg\rfloor+ \dots $$ Note that $3^5=243>190$ and so all $k>4$ terms are zero after taking the floor function of each. Further doing the computational work we have: $$ \mu_3=\bigg\lfloor 63.333... \bigg\rfloor+\bigg\lfloor21.111... \bigg\rfloor+\bigg\lfloor7.037...\bigg\rfloor+\bigg\lfloor2.346...\bigg\rfloor = 63+21+7+2 = 93 $$ which means that $3^{93}$ divides $190!$ by the DePolignac/Legendre formula.

Hopefully it is clear from here how to determine how many 18s divide into $190!$.

Because $18=2\cdot 3^2$ it means that $3^{93}=(3^2)^{46}\cdot 3$ can only provide enough 3s to accommodate $18^{46}$. So the only question is whether $\mu_2>46$ is also true; which could be done by calculating $\mu_2$ but in this case simply knowing $\mu_2>\mu_3$ will suffice to show that $\mu_2>93>46$.

DotCounter
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  • So how do we turn back to 18 as a divisor, from your result? I used a different approach and by shear luck got $18^{30}$, which works. But it is a different number than yours – Luthier415Hz Jul 18 '23 at 15:05
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    @Luthier415Hz I added a spoiler to show the final step of going from the prime factors back to the composite number 18. Does it make sense now? – DotCounter Jul 18 '23 at 15:10
  • Great, thanks...! – Luthier415Hz Jul 18 '23 at 15:12
  • https://www.wolframalpha.com/input?i=190%21%2F18%5E46 . 46 is the right number for 18 as root number. 93 is not the right number for 18 as root. https://www.wolframalpha.com/input?i=190%21%2F18%5E93 – Luthier415Hz Jul 18 '23 at 15:14
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    @Luthier415Hz Correct. The 93 exponent represents the number of 3s that can divide into 190! properly. But each 18 needs two 3s... so $(3^2)^{46}$ divides 190! properly but $(3^2)^{47}$ wouldn't. Reread the part in the spoiler to see if it clicks. – DotCounter Jul 18 '23 at 15:24