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Where $m,n \in \mathbb{N}^{+}$ and $54!= m\cdot 15^{n}$, what is the sum of all possible values of $n$?

I tried to solve it like this:

$$54/ 5 = 10$$ $$10/5=2$$ $$10+2=12$$

Since amount of fives are less than threes in $54!$ I found the total number of fives. Since there are 12 fives, there can be a maximum of 12 fifteens. So highest value of $n$ should be 12. But how do I find others? I thought all numbers smaller than 12 can be applied but the answer turned out to be wrong that way.

EDIT: I contacted the publisher and they confirmed that the answer key is wrong. My initial solution was correct.

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    What should be the answer? – 2017-01-02

2 Answers 2

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You are correct that any $n$ from $0$ to $12$ will work. Without seeing your answer we can't say what is wrong.

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    Well, they are told that $n$ should be positive, so $n$ is from $1$ to $12$. But adding $0$ doesn't change anything. – 2017-01-02
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You get a factor $5$ from $5$, $10$, $15$, $20$, $25$ (two), $30$, $35$, $40$, $45$, $50$, for a total of $12$. The number of factors $3$ is definitely larger. So, in order that $54!$ is divisible by $15^n$ it is necessary and sufficient that $0\le n\le 12$. So the sum of the values is $$ 0+1+2+\dots+12=\frac{12\cdot 13}{2}=78 $$ Whether you count $0$ or not is immaterial.