Find the number of trailing zeros in 50!

Question

My attempt:

50! = 50 * 49 *48 ....

Even * even = even number

Even * odd = even number

odd * odd = odd number

25 evens and 25 odds

Atleast 26 of the numbers will lead to an even multiple (24 evens + 1 even * 1 odd) so at most 26 trailing zeros.

50 is divisible by 5: 10 times. Atleast 10 trailing zeros.

What is the answer?

Isn't dividing by 5 taking into the 25's because it is a multiple? — user9995331, Aug 10 '18 at 15:13
It is. But $25$'s give us two $5$'s. If the factorial was big enough, we would also check for $125$'s, or any powers of $5$ — Jakobian, Aug 10 '18 at 15:14

score 3 · Answer 1 · answered Aug 10 '18 at 15:14

3

Every factor of 5 will mate with a factor of 2 and produce a trailing 0. How do you count the factors of 5 in the product?

answered Aug 10 '18 at 15:14

ncmathsadist

49,383

score 3 · Answer 2 · answered Aug 10 '18 at 15:16

You need to find the highest power of $10$ that divides $50!$, which is same as the highest power of $5$ that divides $50!$, since $10 = 5 \times 2$, and there are fewer multiples of $5$ in the first $50$ numbers. It's easy to count the 10 multiples of $5$, namely $5, 10, 15, \cdots, 50$, but also add two more powers of $5$ as $25$ and $50$ have $5^2$ as the highest power of $5$ as factor.

The answer is $12$.

score 1 · Answer 3 · answered Aug 10 '18 at 15:18

There are $\lfloor \frac{50}{5}\rfloor$ numbers between $1$ and $50$ that are divisible by $5$. Similiarily $\lfloor \frac{50}{5^2}\rfloor$ and $\lfloor \frac{50}{5^3}\rfloor$ numbers divisible by $5^2$ and $5^3$ respectively.

Thus, the highest power of $5$ dividing $50!$ is $$\lfloor \frac{50}{5}\rfloor + \lfloor \frac{50}{5^2}\rfloor + \lfloor \frac{50}{5^3}\rfloor = 10 + 2 + 0 = 12$$

With similiar arguments, one can show the greatest power of $2$ dividing $50!$ is greater then $12$. Thus, the greatest power of $10 = 2 \cdot 5$ dividing $50!$ is $12$.

Find the number of trailing zeros in 50!

3 Answers3