"Whole Factors" of 60! that produce multiples of 10 are:
60, 50, 40, 30, 20, 10
and
2 x 55, 4 x 45, 6 x 35, 8 x 25, 12 x 15, 14 x 5
This totals 12 in number
Surely any other factors are intrinsic to these?
Or indeed, even numbers greater than this cannot be counted as all the multiples of 5 have been paired up. WE CANNOT USE THE NUMBERS TWICE!
Could someone please identify the extra 2 multiples of 10 that produce the final zeros?
Thanks, as I cannot see them.
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Hi Mathsbuddy,
In these types of questions, when the answers are so "close" together, you really have to be thorough with your thinking and your work. You named all of the multiples of 5 and 10 that exist in 60!, but is that REALLY what the question is asking for?
Take the number 50 and multiply it by ANY even number. How many zeroes will you end up with (hint: it's NOT 1).
Now take the number 25. If you multiply it by 2, then you get one 0. What happens when you multiply it by 4 though?
The reason why those extra zeros exist is because both 25 and 50 include two 5s within their respective prime factorizations.
GMAT assassins aren't born, they're made,
Rich
In these types of questions, when the answers are so "close" together, you really have to be thorough with your thinking and your work. You named all of the multiples of 5 and 10 that exist in 60!, but is that REALLY what the question is asking for?
Take the number 50 and multiply it by ANY even number. How many zeroes will you end up with (hint: it's NOT 1).
Now take the number 25. If you multiply it by 2, then you get one 0. What happens when you multiply it by 4 though?
The reason why those extra zeros exist is because both 25 and 50 include two 5s within their respective prime factorizations.
GMAT assassins aren't born, they're made,
Rich
You can solve any of these problems by counting the multiples of 5.
For every multiple of 5, you'll have one zero.
For every multiple of 25, you'll have ONE MORE zero.
For every multiple of 125, you'll have ONE MORE zero, etc.
So for 60!, we have TWELVE multiples of 5 and TWO multiples of 25, giving us 12 + 2 = 14 zeros.
You can generalize this to similar problems. For instance, if they asked how many factors of 3 are found in 60!, we have TWENTY multiples of 3 and SIX multiples of 9 and TWO multiples of 27, giving us 20 + 6 + 2 = 28 factors of 3.
For every multiple of 5, you'll have one zero.
For every multiple of 25, you'll have ONE MORE zero.
For every multiple of 125, you'll have ONE MORE zero, etc.
So for 60!, we have TWELVE multiples of 5 and TWO multiples of 25, giving us 12 + 2 = 14 zeros.
You can generalize this to similar problems. For instance, if they asked how many factors of 3 are found in 60!, we have TWENTY multiples of 3 and SIX multiples of 9 and TWO multiples of 27, giving us 20 + 6 + 2 = 28 factors of 3.
Practice practice practice! Technology is all computers anyway, and the stronger a conceptual problem solver you are the harder you'll be to automate and replace!Kourtney wrote:So hard for me. I think I should back to technology rather than BEAT the GMAT. LOL =))
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To determine the number of trailing zeros in a number, we need to determine the number of 5- and-2 pairs within the prime factorization of that number.Ramit88 wrote:how many trailing zeroes does 60! have?
a 12
b 9
c 15
d 14
e 13
Since we know there are fewer 5s than 2s in 60!, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.
To determine the number of 5s in 60!, we can use the following shortcut in which we divide 60 by 5 and then divide the quotient of 60/5 by 5 and continue this process until we no longer get a nonzero quotient.
60/5 = 12
12/5 = 2 (we can ignore the remainder)
Since 2/5 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 5 within 60!.
Thus, there are 12 + 2 = 14 factors of 5 within 60!, and thus there will be fourteen 5-and-2 pairs, indicating that there are 14 trailing zeros.
Answer: D
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