Trailing zeroes?

[yo4561]

Happy Saturday!

I was watching a Free GMAT Prep Hour with Whitney and she does the following problem:

m=(2^16)(3^15)(4^11)(5^19)
n=(2^19)(3^12)(4^18)(5^16)
If X is the number of trailing zeroes" at the end of M, and Y is the number of "trailing zeroes" at the end of N, what is X-Y?

Solving-->
She then does 19-16=3. Is the general rule that you can take the exponents if the base is 5? I tried this out on other numbers that did not have 5's in the base, and the number of trailing zeroes was not the same.

On a related note, if the number is not in exponent form...
Let's say 270! --> I believe there is a formula online where you divide 270/5 + 270/(5^2) + 270/(5^3) ... how would you know how to do this (if this is correct...do you need to keep going to 270/(5^4) etc.)? But, then again, if you have 21!, then you can't do this when I tried it on a calculator. So, I am guessing the number needs to be divisible by 5?

Overall, what's a smart approach to dealing with trailing zeroes?

[esledge]

I was watching a Free GMAT Prep Hour with Whitney and she does the following problem:

m=(2^16)(3^15)(4^11)(5^19)
n=(2^19)(3^12)(4^18)(5^16)
If X is the number of trailing zeroes" at the end of M, and Y is the number of "trailing zeroes" at the end of N, what is X-Y?

Solving-->
She then does 19-16=3. Is the general rule that you can take the exponents if the base is 5? I tried this out on other numbers that did not have 5's in the base, and the number of trailing zeroes was not the same.

First of all, "trailing zeros" is one of those things where I know what you mean, but it's somewhat ambiguously defined (for example, does 240.00 have 1 trailing zero because 24*10, or 2 trailing zeros because of the decimal places, or is it all 3? Answers I found vary.) And I've never seen the GMAT use this term; I think they would be more likely to describe what they mean without naming it, e.g. "the number of zeros in a row at the end of the integer")

Each zero at the end is due to a power of 10, and each 10 = 2*5. So for that particular problem, the rephrase is really "how many pairs of 2's and 5's in each number?" Whit only used the number of 5's because there were plenty of 2's, so 5's were the limiting factor for making those pairs. Don't memorize this as just looking for 5's in general.

On a related note, if the number is not in exponent form...
Let's say 270! --> I believe there is a formula online where you divide 270/5 + 270/(5^2) + 270/(5^3) ... how would you know how to do this (if this is correct...do you need to keep going to 270/(5^4) etc.)? But, then again, if you have 21!, then you can't do this when I tried it on a calculator. So, I am guessing the number needs to be divisible by 5?

Overall, what's a smart approach to dealing with trailing zeroes?

I don't recognize that formula (though it looks similar to some series expansions, which are way beyond the scope of the GMAT). If you see big factorials on the GMAT, you would only have to recognize some quality they have; you would never have to estimate or compute.

For example, since 270! = 270 * 269 * 268....*2*1 you can recognize that 270! is a multiple of every integer less than or equal to 270. Is it even? Yes. Is it divisible by 187? Yes.

Overall, "trailing zeros" are not really a thing. On the factorial, just understand how a factorial is defined and be prepared to make some inferences accordingly.