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# What is the positive integer n? (1) For every integer m,

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What is the positive integer n? (1) For every integer m, [#permalink]

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18 Aug 2008, 07:32
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What is the positive integer n?

(1) For every integer m, $$(m+n)!/(m-1)!$$ is divisible by 16

(2) $$m^2-9n+20=0$$

Thanks

Kudos [?]: 978 [0], given: 1

Manager
Joined: 04 Jun 2008
Posts: 155

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Re: DS: What is n? [#permalink]

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18 Aug 2008, 07:51
tarek99 wrote:
What is the positive integer n?

(1) For every integer m, $$(m+n)!/(m-1)!$$ is divisible by 16

(2) $$m^2-9n+20=0$$

Thanks

B
back solving

n= (m^2+20) / 9

for n to be a positive integer m^2+20 should be multiple of 9

Try values 9, 18,27 , 36 for m^2+20
m^2+20 = 36 suits best
=> m=+4 or -4

and also n=4;

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SVP
Joined: 30 Apr 2008
Posts: 1870

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Location: Oklahoma City
Schools: Hard Knocks
Re: DS: What is n? [#permalink]

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18 Aug 2008, 07:59
Can you please explain why A is insufficient?

chan4312 wrote:
tarek99 wrote:
What is the positive integer n?

(1) For every integer m, $$(m+n)!/(m-1)!$$ is divisible by 16

(2) $$m^2-9n+20=0$$

Thanks

B
back solving

n= (m^2+20) / 9

for n to be a positive integer m^2+20 should be multiple of 9

Try values 9, 18,27 , 36 for m^2+20
m^2+20 = 36 suits best
=> m=+4 or -4

and also n=4;

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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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Kudos [?]: 607 [0], given: 32

Senior Manager
Joined: 19 Mar 2008
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Re: DS: What is n? [#permalink]

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18 Aug 2008, 08:08
I think the answer is (E).

Because there is upper limit for n. If we insert an super high value for n, the term can surely be divided by 16.

Most likely, I am wrong. I am now in the valley for Quan. Have been too focus on prastising Verb.

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Manager
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Re: DS: What is n? [#permalink]

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18 Aug 2008, 08:29
jallenmorris wrote:
Can you please explain why A is insufficient?

chan4312 wrote:
tarek99 wrote:
What is the positive integer n?

(1) For every integer m, $$(m+n)!/(m-1)!$$ is divisible by 16

(2) $$m^2-9n+20=0$$

Thanks

B
back solving

n= (m^2+20) / 9

for n to be a positive integer m^2+20 should be multiple of 9

Try values 9, 18,27 , 36 for m^2+20
m^2+20 = 36 suits best
=> m=+4 or -4

and also n=4;

put n=1
==> (m+1)! / (m-1)! = (m+1)m

put m=15 ...divisible by 16

put n=2
==> (m+2)! / (m-1)! =(m+2)(m+1)m
put m=14 divisible by 16

put n=3
==> (m+3)! / (m-1)! = (m+3)(m+2)(m+1)m
put m=13

put n=4
==> (m+4)! / (m-1)! = (m+4)(m+3)(m+2)(m+1)m

n can be anything..

So i could not deduce anything from A.

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SVP
Joined: 21 Jul 2006
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Kudos [?]: 978 [0], given: 1

Re: DS: What is n? [#permalink]

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18 Aug 2008, 13:26
Well guys, the OA is C. Here is the OE, which I never understood clearly. Maybe someone could use that to come up with a better explanation:

(1) Note that (m+n)!/(n-1)! = m(m+1)(m+2).....(m+n), the product of the n+1 integers from m to m+n, inclusive. The product is a multiple of 16 for every integer n if and only if three of the terms are even, i.e. if the product consists of at least 6 terms. Thus n+1 is greater than or equal to 6, and n is at least 5. NOT SUFF.

(2) n is either 4 or 5 NOT SUFF.

(1&2) n=5 SUFF.

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Re: DS: What is n? [#permalink]

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18 Aug 2008, 13:43
tarek99 wrote:
Well guys, the OA is C. Here is the OE, which I never understood clearly. Maybe someone could use that to come up with a better explanation:

(1) Note that (m+n)!/(n-1)! = m(m+1)(m+2).....(m+n), the product of the n+1 integers from m to m+n, inclusive. The product is a multiple of 16 for every integer n if and only if three of the terms are even, i.e. if the product consists of at least 6 terms. Thus n+1 is greater than or equal to 6, and n is at least 5. NOT SUFF.

(2) n is either 4 or 5 NOT SUFF.

(1&2) n=5 SUFF.

We can do that!
Let us extend our earlier analysis

n= (m^2+20) / 9

for n to be a positive integer m^2+20 should be multiple of 9

Try values 9, 18,27 , 36, 45 for m^2+20
m^2+20 = 36 => m=+4 or -4
m^2+20 = 45 => m=+5 or -5

substitute m value to get n
==> n can be 4 or 5
let us see which value suits best.

(m+n)!/(n-1)! = m(m+1)(m+2).....(m+n)
for n=4 => it becomes m(m+1)(m+2)(m+3)(m+4)

for a product of consecutive numbers series to have 16 as a factor , there should be atleast three even numbers in the series.
let us see 2*3*4*5*6 = 720 ,divisible by 16

so above n=4 does not suffice.

try n=5 ==> it contains 6 terms..
m can be either even or odd it does not matter..it is divisible by 16.
truly tough one..

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Intern
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Re: DS: What is n? [#permalink]

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18 Aug 2008, 15:59
What is the positive integer n?

(1) For every integer m, is divisible by 16

(2)

Explanation:

From choice 1, we can deduce that n should atleast be 5 because, the numeration will only yield a product of 16, it (m+n) is 6!. This is because, 6! = 6*5*4*3*2*1 and this is the smallest number that has 4 multiples of 2s, which will give 16. So, n>=5. However, this is insufficient.

Choice 2: Here the equation should be n^2-9n+20 = 0 ; The question cannot relate choice 1 and choice 2. There is nowhere else about "m" except in choice 1. Hence, when we solve for the above equation, we get, (n-5)*(n-4)=0, which will give n = 5 or n = 4; So, this is insufficient.

Combining both the choices, we get n = 5 and hence the correct answer is C.

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Re: DS: What is n? [#permalink]

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18 Aug 2008, 19:53
for a product of consecutive numbers series to have 16 as a factor , there should be atleast three even numbers in the series.
let us see 2*3*4*5*6 = 720 ,divisible by 16

.........a very simple question ...........u got the core right

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Re: DS: What is n?   [#permalink] 18 Aug 2008, 19:53
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