salphonso wrote:
Could someone please tell me if this line of thinking is correct?
250*(5!)^2
Here 250 = 5*5*5*2 = 3 trailing zeros because 5 appears 3 times
and (5!)^2 has 2 trailing zeros, which brings the total = 3+2 = 5 trailing zeros.
The number 250 has just one "trailing zero", since there is just one zero at the end of the number. In general, if you can write a number as (2^a)(5^b)(other primes not equal to 2 or 5), then the number of trailing zeros will be the smaller of the two exponents a and b.
And I think when you're considering the number of zeros at the end of the product (250)(5!)^2, you're adding the number of zeros at the end of 250 to the number of zeros at the end of (5!)^2. That won't generally give you the correct answer (it will sometimes, but not always, and not here). For example, the numbers 4 and 25 don't have any trailing zeros, but 4*25 = 100 has two. You genuinely need to find how many 2s and how many 5s are in the product, and use the method I mention above.
All of that said, the question in this thread is either at the very upper end of GMAT difficulty, or is beyond the scope of the test. If any of the relevant concepts are fairly new to you, I would not recommend even attempting this question right now, and I'd generally only suggest spending time on questions like this one if you've mastered all of the commonly tested material, since a question like this one is extremely unlikely to appear on your actual test.