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

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Joined: 12 May 2014
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Re: If n = 20! + 17, then n is divisible by which of the  [#permalink]

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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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31 May 2016, 05:58
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.

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

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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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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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04 Dec 2016, 10:32
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..

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

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18 Nov 2017, 01:18
Bunuel 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.

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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18 Nov 2017, 01:26
3
dave13 wrote:
Bunuel 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.

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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18 Nov 2017, 03:08
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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Re: If n = 20! + 17, then n is divisible by which of the &nbs [#permalink] 18 Nov 2017, 03:08

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