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

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

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New post 11 Sep 2014, 04:13
Take out 17 from 20! and from 17 i.e. 20!+17 = 17 (20*19*18*16*15....*1 + 1), lets say 17 (A+1)

We get, the number is divisible by 17
Now we have two factors 17 & (A+1), clearly 17 is not div by 15 or 19. Also, A is div by 15 as well as 19 hence (A+1) cannot be divisible by 15 or 19.


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

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New post 31 May 2016, 05:58
1
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  [#permalink]

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New post 09 Jun 2016, 22:08
Lets rephrase the question.Putting n=5!+3, then n is divisible by which of the following.
2,3,4

Soln: n=5*4*3*2*1+3=120+3=123
we can take it from here that only 3 can completely divide 123 i.e,n= 5!+3.
Going back to the real figures given and comparing n=x!+y where y =17 in the given question.Hence we can zero in on option .C
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Re: If n = 20! + 17, then n is divisible by which of the  [#permalink]

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New post 03 Dec 2016, 19:20
There are a lot of ways to do this Question.
Here is how i did it ->
Let n=19!+17
clearly 19! is divisible by all integers from 1 to 19 inclusive
n=19k+17 (leaves a remainder 17 with 19)
n=17k(leaves no remainder with 17)
n=15k+2 (leaves a remainder 2 with 15)
Hence B

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

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New post 04 Dec 2016, 10:32
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

\(n = 20! + 17\)

Or, \(n = 20*19*18*17 + 17\)

Or, \(n = 17 ( 20*19*18 + 1 )\)

Or, \(n = 17 ( 6840 + 1 )\)

Or, \(n = 17 * 6841\)

6841 is a prime number so among the given numbers only II satisfy..

Answer will be (C)

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

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New post 18 Nov 2017, 01:18
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  [#permalink]

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New post 18 Nov 2017, 01:26
3
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  [#permalink]

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New post 18 Nov 2017, 03:08
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

C

20! Has 1,2,3, ...20 as factors for sure

Now limiting part is 17

17 is prime

So only 17 can divide it

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

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New post 30 Jan 2019, 05:22
Why so serious?
\(n= 20! + 17\)
Simple divison property :\(\frac{a+b}{c}\) = \(\frac{a}{c}\) +\(\frac{b}{c}\)
Therefore lets plug in all the numbers
1. 15 :\(\frac{20!}{15}\) + \(\frac{17}{15}\) not an integer
2. 17 :\(\frac{20!}{17}\) +\(\frac{17}{17}\) is an integer
3. 19 :\(\frac{20!}{19}\)+ \(\frac{17}{19}\) is not an integer
Hence Answer C.
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Re: If n = 20! + 17, then n is divisible by which of the  [#permalink]

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New post 15 Feb 2019, 09:29
Top Contributor
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  [#permalink]

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New post 15 Feb 2019, 10:39
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


Both 20! and 17 have 17 in common, therefore n is divisible by 17
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Re: If n = 20! + 17, then n is divisible by which of the   [#permalink] 15 Feb 2019, 10:39

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