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If n = 20! + 17, then n is divisible by which of the following?

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If n = 20! + 17, then n is divisible by which of the following?  [#permalink]

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New post 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 III
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Re: If n = 20! + 17, then n is divisible by which of the following?  [#permalink]

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New post 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 \(a-b=-3\), 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 \(a-b=1\), 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 \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), 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 following?  [#permalink]

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New post 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 following?  [#permalink]

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New post 07 Jun 2013, 02:52
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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


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 ... c-or-math/

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 following?  [#permalink]

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New post 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 following?  [#permalink]

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New post 16 Jun 2013, 06:19
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Re: If n = 20! + 17, then n is divisible by which of the following?  [#permalink]

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New post 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 following?  [#permalink]

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New post 25 Sep 2013, 21:20
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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 following?  [#permalink]

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New post 22 Oct 2013, 13:33
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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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Re: If n = 20! + 17, then n is divisible by which of the following?  [#permalink]

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New post 22 Oct 2013, 20:50
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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 following?  [#permalink]

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New post 31 May 2016, 06:58
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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


We are given that n = 20! + 17 and need to know whether n is divisible by 15, 17, and/or 19. To determine this, we rewrite the given expression for n using each answer choice.

Thus, we have:

Does (20! + 17)/15 = integer?

Does (20! + 17)/17 = integer?

Does (20! + 17)/19 = integer?

We now use the distributive property of division over addition to determine which of these expressions is/are equal to an integer.

The distributive property of division over addition tells us that (a + c)/b = a/b + c/b. We apply this rule as follows:

I.

Does (20! + 17)/15 = integer?

Does 20!/15 + 17/15 = integer?

Although 20! is divisible by 15, 17 is NOT, and thus (20! + 17)/15 IS NOT an integer.

We can eliminate answer choices B and D.

II.

Does (20! + 17)/17 = integer?

Does 20!/17 + 17/17 = integer?

Both 20! and 17 are divisible by 17, and thus (20! + 17)/17 IS an integer.

We can eliminate answer choice A.

III.

Does (20! + 17)/19 = integer?

Does 20!/19 + 17/19 = integer?

Although 20! is divisible by 19, 17 is NOT, so (20! + 17)/19 IS NOT an integer.

We can eliminate answer choice E.

Thus, II is the only correct statement.

Answer: C
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Re: If n = 20! + 17, then n is divisible by which of the following?  [#permalink]

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New post 18 Nov 2017, 02:18
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 \(a-b=-3\), 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 \(a-b=1\), 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 \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope it helps.


Bunuel, Greetings! :) What is the point of taking "17" out of brackets? what is the point of factoring here? How do we get "1" ? Also I am still trying to understand if 20! includes both "15" and 19" than why 20!+17 isn't divisible by "15" and 19". So what if there are two "17" s, ....."15" and 19" are still in 20! :? thanks for taking time to explain and have a great day:)
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Re: If n = 20! + 17, then n is divisible by which of the following?  [#permalink]

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New post 18 Nov 2017, 02:26
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dave13 wrote:
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 \(a-b=-3\), 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 \(a-b=1\), 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 \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope it helps.


Bunuel, Greetings! :) What is the point of taking "17" out of brackets? what is the point of factoring here? How do we get "1" ? Also I am still trying to understand if 20! includes both "15" and 19" than why 20!+17 isn't divisible by "15" and 19". So what if there are two "17" s, ....."15" and 19" are still in 20! :? thanks for taking time to explain and have a great day:)



20! + 17 = 17*(1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20 + 1), so it's divisible by 17.

Next, pleas re-read the highlighted part carefully.

20! IS divisible by 15 but 17 is NOT. Thus 20! + 17 = (a multiple of 15) + (NOT a multiple of 15) = (NOT a multiple of 15);
20! IS divisible by 19 but 17 is NOT. Thus 20! + 17 = (a multiple of 19) + (NOT a multiple of 19) = (NOT a multiple of 19).
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Re: If n = 20! + 17, then n is divisible by which of the following?  [#permalink]

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New post 15 Feb 2019, 10:29
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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 III



Answer choice I: is 20! + 17 divisible by 15?
20! + 17 = (20)(19)(18)(17)(16)(15)(other stuff) + 15 + 2
= (15)(some number + 1) + 2
(15)(some number + 1) is a multiple of 15
So, (15)(some number + 1) + 2 is 2 greater than a multiple of 15
So, if we divide (15)(some number + 1) + 2 by 15, the remainder will be 2
So, 20! + 17 is NOT divisible by 15
ELIMINATE B and D

Answer choice II: is 20! + 17 divisible by 17?
20! + 17 = (20)(19)(18)(17)(other stuff) + 17
= (17)(some number + 1)
If we divide (17)(some number + 1) by 17, the remainder will be 0
So, 20! + 17 IS divisible by 17
ELIMINATE A

Answer choice III: is 20! + 17 divisible by 19?
20! + 17 = (20)(19)(other stuff) + 17
= (19)(some number) + 17
If we divide (19)(some number) + 17 by 19, the remainder will be 17
So, 20! + 17 is NOT divisible by 19
ELIMINATE E


Answer: C

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Re: If n = 20! + 17, then n is divisible by which of the following?  [#permalink]

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Re: If n = 20! + 17, then n is divisible by which of the following?   [#permalink] 30 Jul 2019, 01:41
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