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If n = 20! + 17, then n is divisible by which of the [#permalink]
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05 Dec 2012, 09:02
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If n = 20! + 17, then n is divisible by which of the following? I. 15 II. 17 III. 19 (A) None (B) I only (C) II only (D) I and II (E) II and II
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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05 Dec 2012, 09:06
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Walkabout wrote: If n = 20! + 17, then n is divisible by which of the following?
I. 15 II. 17 III. 19
(A) None (B) I only (C) II only (D) I and II (E) II and II 20! is the product of all integers from 1 to 20, inclusive, thus it's divisible by each of the integers 15, 17, and 19. Next, notice that we can factor out 17 from from 20! + 17, thus 20! + 17 is divisible by 17 but we cannot factor out neither 15 nor 19 from 20! + 17, thus 20! + 17 is not divisible by either of them. Answer: C. GENERALLY: If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\) and \(b=9\), both divisible by 3 > \(a+b=15\) and \(ab=3\), again both divisible by 3. If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 > \(a+b=11\) and \(ab=1\), neither is divisible by 3. If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):Example: \(a=5\) and \(b=4\), neither is divisible by 3 > \(a+b=9\), is divisible by 3 and \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), both are divisible by 4. Hope it helps.
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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05 Dec 2012, 09:12
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Here its how it works we are given 20*19*18....*1 + 17
option 1 15 now if we try to divide by 15 the first part is 20!/15 ( this part is divisible) + 17/15 ( this isn't)
option 2 17 we can take a 17 common and then its divisible
option 3 the first part is 20!/19 ( this part is divisible) + 17/19 ( not divisble)
hence option C.



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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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07 Jun 2013, 02:52
Walkabout wrote: If n = 20! + 17, then n is divisible by which of the following?
I. 15 II. 17 III. 19
(A) None (B) I only (C) II only (D) I and II (E) II and II Another point to note here is that if a number n is divisible by m (where m is greater than 1), (n+1) will not be divisible by m. The logic is explained here: http://www.veritasprep.com/blog/2011/09 ... cormath/So 20! + 17 = 17*(1*2*3*...15*16*18*19*20 + 1) Assume 1*2*3*...15*16*18*19*20 = N So 20! + 17 = 17(N + 1) If this number has to be divisible by either 15 or 19, (N+ 1) must be divisible by 15 or 19. Since N is divisible by both 15 and 19, (N + 1) can be divisible by neither.
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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18 Apr 2013, 23:07
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Another crude way to answer this, if you did not know the properties above would be to consider that that 20! will have the number ending in 00 due to 10 and 20 being included.
So n!+17 = xxxx00 +17 = xxxx17 which is only possibly divisible by 17. Hence Option C is the answer.
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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16 Jun 2013, 06:19



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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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25 Sep 2013, 21:20
TAL010 wrote: Is this type of problem similar to remainder problems? My thinking was that n = 20(n) + 17 with zero remainder, meaning it had to be divisible by 17. If it were divisible by 17, then it was not divisible by 15 or 19. Does that make sense though? I am not sure I understand what you did there. How do you get n = 20a + 17 has to be divisible by 17? (I am assuming you meant the second variable to be different) The reason 20! + 17 must be divisible by 17 is that 20! = 1*2*3*4*...17*18*19*20 So 20! is a multiple of 17 and can be written as 17a n = 17a + 17 is divisible by 17. Also, a number can be divisible by 17 as well as 15 as well as 19. e.g. n =15*17*19 is divisible by all three. But here n = 20! + 17 in which 20! is divisible by 15 as well as 19 but 17 is neither divisible by 15 nor by 19. So when you divide n by 15, you will get a remainder of 2. When you divide n by 19, you will get a remainder of 17.
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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22 Oct 2013, 20:50
TAL010 wrote: I was thinking that n = 20! + 17  knowing this you know that 20! is a multiple of all of the numbers 15, 17 and 19, because 20! = 20x19x18x17x16x15...etc.etc.
However by adding 17, the number is no longer a multiple of 15 or 19 because 17 is not divisible by 19 or 15.
Yes, this is fine. Note that previously, you had written n = 20a + 17 is divisible by 17. I am assuming the 'a' was a typo.
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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18 Apr 2013, 23:51
mal208213 wrote: Another crude way to answer this, if you did not know the properties above would be to consider that that 20! will have the number ending in 00 due to 10 and 20 being included.
So n!+17 = xxxx00 +17 = xxxx17 which is only possibly divisible by 17. Hence Option C is the answer.
Please give Kudos if you found this useful! Not true. 19*43 = _17. There are bigger numbers ending with 17 and are divisible by 19. By your logic, you can only eliminate 15.
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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19 Apr 2013, 00:02
mal208213 wrote: Another crude way to answer this, if you did not know the properties above would be to consider that that 20! will have the number ending in 00 due to 10 and 20 being included.
So n!+17 = xxxx00 +17 = xxxx17 which is only possibly divisible by 17. Hence Option C is the answer.
Please give Kudos if you found this useful! This is not true mate. Consider \(16!\), it has two \(00\) at the end (\(3\) if we want to be precise) but \(16! +17\) => \(xxx17\) is NOT divisible by \(17\) In order for an \(n!\) to be divisible by \(17\) we have to be able to group/take 17 from both numbers \(17!+34=17(16!+2)\) Divisible by 17 \(17!+35\) Not divisible by 17 Hope this clarifies the concept, let me know
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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19 Apr 2013, 10:48
Zarrolou wrote: mal208213 wrote: Another crude way to answer this, if you did not know the properties above would be to consider that that 20! will have the number ending in 00 due to 10 and 20 being included.
So n!+17 = xxxx00 +17 = xxxx17 which is only possibly divisible by 17. Hence Option C is the answer.
Please give Kudos if you found this useful! This is not true mate. Consider \(16!\), it has two \(00\) at the end (\(3\) if we want to be precise) but \(16! +17\) => \(xxx17\) is NOT divisible by \(17\) In order for an \(n!\) to be divisible by \(17\) we have to be able to group/take 17 from both numbers \(17!+34=17(16!+2)\) Divisible by 17 \(17!+35\) Not divisible by 17 Hope this clarifies the concept, let me know Sure Zarroulou and Vinay that definitely clarifies.... Again my method was only in the moment when I found myself taking way too much time to solve this and my solution was at best an estimated guess or "guesstimate"



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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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06 Jun 2013, 17:33
thks for posting the question.



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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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16 Jun 2013, 06:15
But in this case !20+17 is not completely divisible by 17.So 17 is not the factor of sum of !20+17.Please clarify?



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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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16 Jun 2013, 06:55
Bunuel wrote: anu1706 wrote: But in this case !20+17 is not completely divisible by 17.So 17 is not the factor of sum of !20+17.Please clarify? Actually it is: \(20! + 17=17(1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20+1)\) \(\frac{20! + 17}{17}=1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20+1\) Hope it's clear. I totally understood the concept Bunuel!!Thanks a ton for your contribution, but my point is that I got confused with the wordings of the question stem. If it says n is divisible by 17, so that means it should be divisible completely which is not case in this question.So how to check in a question like this that what concept to apply like completely divisible or one of the factor is divisible? Please clarify.Hope you got my point.



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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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16 Jun 2013, 07:00
Bunuel wrote: Walkabout wrote: If n = 20! + 17, then n is divisible by which of the following?
I. 15 II. 17 III. 19
(A) None (B) I only (C) II only (D) I and II (E) II and II 20! is the product of all integers from 1 to 20, inclusive, thus it's divisible by each of the integers 15, 17, and 19. Next, notice that we can factor out 17 from from 20! + 17, thus 20! + 17 is divisible by 17 but we cannot factor out neither 15 nor 19 from 20! + 17, thus 20! + 17 is not divisible by either of them. Answer: C. GENERALLY: If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\) and \(b=9\), both divisible by 3 > \(a+b=15\) and \(ab=3\), again both divisible by 3. If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 > \(a+b=11\) and \(ab=1\), neither is divisible by 3. If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):Example: \(a=5\) and \(b=4\), neither is divisible by 3 > \(a+b=9\), is divisible by 3 and \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), both are divisible by 4. Hope it helps. Please tell me for the multiplication cases as well like you explained about addition & subtraction.



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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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16 Jun 2013, 08:00
anu1706 wrote: Bunuel wrote: anu1706 wrote: But in this case !20+17 is not completely divisible by 17.So 17 is not the factor of sum of !20+17.Please clarify? Actually it is: \(20! + 17=17(1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20+1)\) \(\frac{20! + 17}{17}=1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20+1\) Hope it's clear. I totally understood the concept Bunuel!!Thanks a ton for your contribution, but my point is that I got confused with the wordings of the question stem. If it says n is divisible by 17, so that means it should be divisible completely which is not case in this question.So how to check in a question like this that what concept to apply like completely divisible or one of the factor is divisible? Please clarify.Hope you got my point. What do you mean "divisible completely"? There is no such thing. 20!+17 is divisible by 17 means that (20!+17)/17=integer, which is the case here.
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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16 Jun 2013, 08:28
By completely divisible I mean to say that Remainder is Zero which in this case it is not!!



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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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25 Sep 2013, 14:07
Is this type of problem similar to remainder problems? My thinking was that n = 20(n) + 17 with zero remainder, meaning it had to be divisible by 17. If it were divisible by 17, then it was not divisible by 15 or 19. Does that make sense though?
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Re: If n = 20! + 17, then n is divisible by which of the [#permalink]
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22 Oct 2013, 13:33
I was thinking that n = 20! + 17  knowing this you know that 20! is a multiple of all of the numbers 15, 17 and 19, because 20! = 20x19x18x17x16x15...etc.etc. However by adding 17, the number is no longer a multiple of 15 or 19 because 17 is not divisible by 19 or 15. i VeritasPrepKarishma wrote: TAL010 wrote: Is this type of problem similar to remainder problems? My thinking was that n = 20(n) + 17 with zero remainder, meaning it had to be divisible by 17. If it were divisible by 17, then it was not divisible by 15 or 19. Does that make sense though? I am not sure I understand what you did there. How do you get n = 20a + 17 has to be divisible by 17? (I am assuming you meant the second variable to be different) The reason 20! + 17 must be divisible by 17 is that 20! = 1*2*3*4*...17*18*19*20 So 20! is a multiple of 17 and can be written as 17a n = 17a + 17 is divisible by 17. Also, a number can be divisible by 17 as well as 15 as well as 19. e.g. n =15*17*19 is divisible by all three. But here n = 20! + 17 in which 20! is divisible by 15 as well as 19 but 17 is neither divisible by 15 nor by 19. So when you divide n by 15, you will get a remainder of 2. When you divide n by 19, you will get a remainder of 17.
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