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If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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Updated on: 12 Nov 2014, 04:27
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If 30!/10! is written as the product of consecutive integers, the largest of which is 30, what is the smallest of the integers? A. 1 B. 3 C. 7 D. 11 E. 20
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Originally posted by MDK on 12 Oct 2013, 04:01.
Last edited by Bunuel on 12 Nov 2014, 04:27, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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12 Oct 2013, 04:03
lucbesson wrote: If \(30!/10!\) is witten as the product of consequtive integers, the largest of which is 30, what is the smallest of the integers? Once you see the answer, the logic becames clear. But can anyone suggest how to deal with such kind of problems? THX!



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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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12 Oct 2013, 04:22
Quite straight forward actually. The answer is 11. 30! = 1*2*3......*29*30 10! = 1*2*3......*9*10 So, 1 to 10 gets cancelled out and the remaining series begins with 11. (1 cannot be considered since the series has to be consecutive integers)
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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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12 Oct 2013, 04:27
MacFauz wrote: Quite straight forward actually. The answer is 11. 30! = 1*2*3......*29*30 10! = 1*2*3......*9*10
So, 1 to 10 gets cancelled out and the remaining series begins with 11. (1 cannot be considered since the series has to be consecutive integers) Yeahhh, you are 100% right! I accidentally thought that the fraction transforms to 20*21*...*30, so I was looking for some very complicated method to extract 11 from 22 and make another very smart sequence ) Probably I was too tired But thanks anyway!!!



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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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11 Nov 2014, 16:49
Bunuel or WoundedTigerI have tagged this question as I faced it in the GMAT Prep Exam pack 1, however, the original post is missing the answer choices, Can you please edit the original post to include the answer choices as follows?
A) 1 B) 3 C) 7 D) 11 E) 20
Thanks! DmmK



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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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12 Nov 2014, 00:10
MDK wrote: If \(30!/10!\) is witten as the product of consequtive integers, the largest of which is 30, what is the smallest of the integers? \(\frac{30!}{10!} = \frac{30 * 29 * 28........... 11 * 10!}{10!}\) Smallest integer = 11 Answer = D
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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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12 Nov 2014, 04:27



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If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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18 Aug 2015, 08:25
MacFauz wrote: Quite straight forward actually. The answer is 11. 30! = 1*2*3......*29*30 10! = 1*2*3......*9*10
So, 1 to 10 gets cancelled out and the remaining series begins with 11. (1 cannot be considered since the series has to be consecutive integers) Hi, Just did the GMAT test prep and this question popped up. Basically, my first answer was the good one, 11, but after double checking, the smallest of the integers, after simplifying the 30*29*28*...*11, is actually 1, because 1*30*29*28... It is possible to breakdown any of the numbers and multiplying them by one. So I went for 1. Anyone could tell me from the question, how could I have decided between 11 and 1? My understanding is that the question specifies "the product of consecutive integer" only to explain what means 30 !Thanks a lot.



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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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29 Aug 2015, 10:31
Hi tsunagaru, You are right that multiplying the sequence by 1 would still give the same answer. However, including 1 in the sequence would mean that the sequence is not a sequence of consecutive integers any more. As you have mentioned, the sequence would then become 1*11*12*13.....29*30 As you can see this would not be a valid sequence. tsunagaru wrote: MacFauz wrote: Quite straight forward actually. The answer is 11. 30! = 1*2*3......*29*30 10! = 1*2*3......*9*10
So, 1 to 10 gets cancelled out and the remaining series begins with 11. (1 cannot be considered since the series has to be consecutive integers) Hi, Just did the GMAT test prep and this question popped up. Basically, my first answer was the good one, 11, but after double checking, the smallest of the integers, after simplifying the 30*29*28*...*11, is actually 1, because 1*30*29*28... It is possible to breakdown any of the numbers and multiplying them by one. So I went for 1. Anyone could tell me from the question, how could I have decided between 11 and 1? My understanding is that the question specifies "the product of consecutive integer" only to explain what means 30 !Thanks a lot.
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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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16 Feb 2017, 08:40
MDK wrote: If 30!/10! is written as the product of consecutive integers, the largest of which is 30, what is the smallest of the integers?
A. 1 B. 3 C. 7 D. 11 E. 20 30!/10! \(= \frac{30*29*28...........13*12*11*10!}{10!}\) Thus we are left with : 30*29*28...........13*12*11 The largest number here is 30 and thus the smallest number will be 11, answer will be (D) 11
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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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21 Feb 2017, 10:21
MDK wrote: If 30!/10! is written as the product of consecutive integers, the largest of which is 30, what is the smallest of the integers?
A. 1 B. 3 C. 7 D. 11 E. 20 We can simplify 30!/10! to 30 x 29 x 28 x 27 x … x 13 x 12 x 11. Thus, the smallest integer is 11. Answer: D
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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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22 Feb 2017, 20:42
Hi All, When a prompt provides you with a 'complexlooking' calculation, it can often help to come up with a much simpler example of the math involved. Here, we're asked to consider 30!/10!  and that might look 'scary' at first glance. Instead of starting with that example, consider the following  what would 5!/3! look like...? (5)(4)(3)(2)(1) / (3)(2)(1) = 120/6 = 20 You probably already know that when you simplify a fraction, you divide "top" and "bottom" by the same number, so we can 'cancel out' the 3, 2 and 1 from both the numerator and the denominator. This leaves us with... (5)(4)/1 = 20 That wasn't too difficult, so now we can apply the same logic to 30!/10!... All of the numbers (10), (9)....(2) and (1) will cancel out, leaving us with... (30)(29)....(12)(11)/1 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If 30!/10! is written as the product of consecutive integers, the larg [#permalink]
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03 Apr 2017, 05:58
30!= 1*2*3*4*5*6*7*8*9*10*11*....................*30 10!= 1*2*3*4*5*6*7*8*9*10Therefore 30!/10!= 11*12*............30 (as all common terms i.e. 10! cancel out). Thus the smallest integer is 11.
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