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I'm trying to figure out the pattern:

There is 10 , 20, 30 whose product gives three 0s (6000)

Then there are the following that would result in 10: 2 x 5 = 10 4 x 15 = 60 6 x 25 = 150

So, initially I thought there are 6 zeros by merely counting the zeros. But the product (6000 * 10 * 60 * 150) was giving me 7 zeros. I realized it is because 60 * 150 gives me another zero (9000)

Rewriting/rearranging seems to make this clear: 2 x 5 = 10 6 x 15 = 90 4 x 25 = 100

So, even though I got 7 0s, I was wondering if there is a pattern in these question or if is trial-error method only. Is there an efficient way to answer this question.

I have the explanation in the book and it seems to be a bit convoluted. I can type in the explanation if someone wants to take a look.

Thanks, grajee

Question: How many identical digits end the product? 1 x 2 x 3 x 4 x 5 x 6 x 7 x .... x 30

Answer: 7

Gopinath Rajee
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3 Answers3

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The most general answer: you should count the number of factors $2$ and $5$ in the product, and the minimum of these two numbers gives you the number of factors $10$. In most cases, it boils down to counting the number of factors $5$ you find in the product.

In your example the answer is $7$ because of the factors $5$ in $5, 10, 15, 20, 25, 25, 30$. Note that $25$ contains two factors $5$.

TMM
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Given positive integers $n_1, \ldots, n_k$, let $a_i$ be the exponent of $2$ in the prime factorization of $n_i$, and $b_i$ be the exponent of $5$ in $n_i$. Then we have the number of zeroes at the end of $n_1 \ldots n_k$ is equal to $\min\{ a_1 + \ldots + a_k, b_1 + \ldots + b_k\}$.

For $1 \cdot 2 \cdot \ldots \cdot 30$ we have 15 terms divisible by $2$, 7 terms divisible by $4$, 3 terms divisible by $8$, and 1 term divisible by $16$. So $a_1 + \ldots + a_{30} = 26$.

We have 6 terms divisible by $5$, and 1 term divisible by $25$, so $b_1 + \ldots + b_{30} = 7$.

The minimum of these is $7$.

In general when looking at the number of zeroes at the ends of $n!$, we can argue its going to be $\sum_{i=1}^{\log_5(n)} \lfloor \frac{n}{5^i} \rfloor$

Nick R
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Note that a $0$ is pretty much just a $5$ and a $2$. In a product like this, it is rather easy to see that we'll have more $2$s than $5$s, so the number of $0$s at the end is just the number of $5$s in the prime factorization.

In general, the number of factors $p$ in $n!$, with $p$ prime, is

$$\sum_{k=1}^{\infty} \left \lfloor\frac{n}{p^k}\right \rfloor $$

(Note that if $p^k>n$, these terms are $0$, so the sum is actually finite). It's a good exercise to prove this.

Carl Schildkraut
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