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If 60! is written out as an integer, with how many

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Re: How many zeroes at the end of 60!? [#permalink]

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New post 19 May 2014, 18:06
VeritasPrepKarishma wrote:
russ9 wrote:
How do we know that there aren't more factors of 2 vs. 5? If there were more factors of 2, would we modify the equation to account for powers of 2 in the denominator?

Thanks!


The point is that you need both a 2 and a 5 to make a 10. If I have 100 2s but only 3 5s, I can make only 3 10s. No number of 2s alone can make a 10. So even if there are many more 2s, they are useless to us because we have limited number of 5s.


Hi Karishma and @WoundedTiger, makes sense. Thanks!

Would it make sense to compare the 2's in the denominator as well and choose the one's that have the lower exponent out of the two or are the 5's ALWAYS going to be less than the 2's?
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Re: How many zeroes at the end of 60!? [#permalink]

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New post 19 May 2014, 19:41
russ9 wrote:
VeritasPrepKarishma wrote:
russ9 wrote:
How do we know that there aren't more factors of 2 vs. 5? If there were more factors of 2, would we modify the equation to account for powers of 2 in the denominator?

Thanks!


The point is that you need both a 2 and a 5 to make a 10. If I have 100 2s but only 3 5s, I can make only 3 10s. No number of 2s alone can make a 10. So even if there are many more 2s, they are useless to us because we have limited number of 5s.


Hi Karishma and @WoundedTiger, makes sense. Thanks!

Would it make sense to compare the 2's in the denominator as well and choose the one's that have the lower exponent out of the two or are the 5's ALWAYS going to be less than the 2's?


Every second number has a 2 while every fifth number has a 5. So number of 2s will always be greater than the number of 5s.
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Re: If 60! is written out as an integer, with how many [#permalink]

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New post 29 Apr 2015, 20:09
Awesome learning from this post ; Thanks everyone
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Re: If 60! is written out as an integer, with how many [#permalink]

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New post 25 Sep 2016, 03:36
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56



Correct answer is C.

60/5 + 60/25. Quotient is 12 + 2 = 14
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Re: If 60! is written out as an integer, with how many [#permalink]

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New post 25 Sep 2016, 08:11
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56


10 Sec Solution

60/5 = 12
12/5 = 2

So, Trailing Zeroe's will be 14 ( 12 + 2 ) , answer will be (C) 14

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Re: If 60! is written out as an integer, with how many [#permalink]

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New post 10 Feb 2017, 15:52
Orange08 wrote:
Many thanks.
silly question probably - can real test contain such question?


Regarding the concept, which Bunuel mentioned, yes. However, I would expect the wording to be different.
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Re: If 60! is written out as an integer, with how many [#permalink]

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New post 22 May 2017, 20:28
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56


Here it's best to think about the problem conceptually. If we want to know then number of consecutive zeroes at the end of 60!, then we are really asking for the number of pairs of 2's and 5's. Since there are more multiples of 2 than 5 within 1-60, 5 will be the limiting factor/agent.

As a result, we can simplify find the number of multiples of 5 between 1 and 60 (12) then add two more for 50 and 25 that each contribute an extra five. The total number of 5's in this prime factorization is thus 12+2 =14.
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Re: If 60! is written out as an integer, with how many [#permalink]

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New post 19 Apr 2018, 16:07
Top Contributor
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56


60! = (1)(2)(3)(4)(5)(6) . . . (57)(58)(59)(60)

KEY CONCEPT: For every pair of one 2 and one 5, we get a product of 10, which accounts for one zero at the end of the integer.
So, the question is "How many pairs of one 2 and one 5 are "hiding" in the product?"

Well, there is no shortage of 2's hiding in the product. In fact, there are FAR MORE 2's than 5's. So, all we need to do is determine how many 5's are hiding in the product.
60! = (1)(2)(3)(4)(5)(6) . . . (57)(58)(59)(60)
= (1)(2)(3)(4)(5)(6)(7)(8)(9)(2)(5)(11)...(3)(5)...(4)(5)...(5)(5)...(6)(5)...(7)(5)...(8)(5)...(9)(5)...(2)(5)(5)...(11)(5)...(56)(57)(58)(59)(12)(5)
In total, there are 14 5's hiding in the product.
And there are MORE THAN 14 2's hiding in the product.

So, there are 14 pairs of 2's and 5'2, which means the integer ends with 14 zeros

Answer:C

Cheers,
Brent
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Re: If 60! is written out as an integer, with how many   [#permalink] 19 Apr 2018, 16:07

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