Number of zeros

Number Theory Level 1

Find the number of trailing zeros in the following expression.

$1!^2 \times 2!^2 \times \cdots \times 30!^2.$

266 174 cannot be determined 88

3 solutions

Vladimir Smith
Dec 15, 2015

Let the given expression be equal to x. When we square root x, we will have half the zeroes we previously had. This amount of zeroes still has to be a whole number of zeroes - we can't have half a zero. Thus, the original number of zeroes in x must be an even amount. We pick the only even option; the answer is 174.

Good application of logic.

tanay gaurav - 5 years, 6 months ago

Thanks. I've updated the options so that they are all even.

Calvin Lin Staff - 5 years, 5 months ago

Tanay Gaurav
Dec 12, 2015

No of zeros in in (1!^2 x 2!^2 x...x 30!^2)=1 5 + 2 5 +3 5 +4 5 +6 5 +7 1 = 87 So number of zeros in the given expression 187*2 = 174

Arulx Z
Dec 21, 2015

This answer is just an edited elaboration of Vladimir Smith's method. Do check his answer out and leave an upvote.

$s=1!\times 2!\times \dots \times 30!$

Let the number of trailing 0s be $n$ .

By experimenting a bit, it can be found that the number of trailing 0s in $s^2$ is $2n$ . For any integral $n$ , $2n$ will always be even. Since 174 is the only even option, it is the answer.

It is useful to look at parities, especially when provided with options.

What if I would have given another even number as option ?

tanay gaurav - 5 years, 5 months ago

Then I would have used this method to solve the problem (with shortcuts of course!). In fact I did it using the trailing zeroes method myself and it is pretty straight forward. But I found Vladimir's proof more innovative. That's why I posted this answer.

Arulx Z - 5 years, 5 months ago

Nice work.