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Re: How many zeroes at the end of 60!?
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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!?
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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
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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
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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
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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 Solution60/5 = 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
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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
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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 160, 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
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19 Apr 2018, 16:07
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
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15 Sep 2018, 12:34
I'm a bit confused with the concept of "Trailing zeros". According to wikipedia, "Trailing zeros to the right of a decimal point, as in 12.3400, do not affect the value of a number and may be omitted if all that is of interest is its numerical value. This is true even if the zeros recur infinitely". However, zeros to the left of a decimal point are introduced as "trailing zeros" in this forum. I just wanted to clarify the meaning of "trailing zeros" here.
Thank you!!



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Re: If 60! is written out as an integer, with how many
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15 Sep 2018, 13:46
owlette wrote: I'm a bit confused with the concept of "Trailing zeros". According to wikipedia, "Trailing zeros to the right of a decimal point, as in 12.3400, do not affect the value of a number and may be omitted if all that is of interest is its numerical value. This is true even if the zeros recur infinitely". However, zeros to the left of a decimal point are introduced as "trailing zeros" in this forum. I just wanted to clarify the meaning of "trailing zeros" here.
Thank you!! It is certain that 60! will not be a decimal but a integer. Anything larger than 5! Ends in a zero. A trailing zero here is the number of zeros to the right of first nonzero digit and to the left of the decimal point. Integer 2 can be written as 2.0 So 60! = ABCxx0000xx0.0 Hope you get the usage. Best, Gladi Posted from my mobile device
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Re: If 60! is written out as an integer, with how many
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16 Sep 2018, 01:21
owlette wrote: I'm a bit confused with the concept of "Trailing zeros". According to wikipedia, "Trailing zeros to the right of a decimal point, as in 12.3400, do not affect the value of a number and may be omitted if all that is of interest is its numerical value. This is true even if the zeros recur infinitely". However, zeros to the left of a decimal point are introduced as "trailing zeros" in this forum. I just wanted to clarify the meaning of "trailing zeros" here.
Thank you!! Links below should help: Everything about Factorials on the GMATPower of a Number in a Factorial ProblemsTrailing Zeros ProblemsFor more check Ultimate GMAT Quantitative MegathreadHope it helps.
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Re: If 60! is written out as an integer, with how many
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15 Mar 2019, 18:45
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 To determine the number of trailing zeros in a number, we need to determine the number of 5and2 pairs within the prime factorization of that number. Since we know there are fewer 5s in 60! than 2s, we can find the number of 5s and thus be able to determine the number of 5and2 pairs. To determine the number of 5s within 60!, we can use the following shortcut in which we divide 60 by 5, 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! Since there are 14 factors of 5 within 60!, there are 14 5and2 pairs and thus 14 trailing zeros. Answer: C
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Re: If 60! is written out as an integer, with how many
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20 Sep 2019, 05:55
Is there any other way to count the number of trailing zeroes? What if we forget the formula in exam  can I calculate by counting 2s*5s and number of 10s (10, 20, 30, 40,..). This way I counted 13 trailing zeroes and not 14. Can someone please explain.



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Re: If 60! is written out as an integer, with how many
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02 Jan 2020, 07:21
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 Let's first make a few observations: 1300 = (2)(2)(5)(5)(13) 8000 = (2)(2)(2)(2)(2)(2)(5)(5)(5) 140,000 = (2)(2)(2)(2)(2)(5)(5)(5)(5)(7) Notice that for each PAIR of one 2 and one 5, we get a zero at the end of the number. So, our question really becomes "How many PAIRS of one 2 and one 5 are "hiding" in the prime factorization of 60!" ASIDE: There are wayyyyyy more 2's hiding in the prime factorization of 60! than there are 5's hiding in the prime factorization. So, if we know the number of 5's hiding in the prime factorization, then we can rest assured knowing that there will be enough 2's to pair up with each 5. So, let's determine how many 5's are "hiding" in the prime factorization of 60! 60! = (60)(59)(58)......(3)(2)(1) 60 = (2)(2)(3)(5) [one 5]55 = (5)(11) [one 5]50 = (2)(5)(5) [two 5's]45 [one 5]40 [one 5]35 [one 5]30 [one 5]25 [two 5's]20 [one 5]15 [one 5]10 [one 5]5 [one 5]Add them all to get: 14 fives Answer: C Cheers, Brent
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Re: If 60! is written out as an integer, with how many
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