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positive integer n

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by das.ashmita » Wed Oct 03, 2012 11:05 am
What is the positive integer n?

(1) For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
(2) n^2 - 9n + 20 = 0

OA in the source is C

But I think its E
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Source: — Data Sufficiency |
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by anuprajan5 » Thu Oct 04, 2012 6:16 am
Ashmita,

The answer is C

Statement 1 - m(m=1)(m=2)....(m+n) is divisible by 16. Since these are consecutive integers, I only need a couple of even integers to make it divisible by 16.

For example, 1,2,3,4,5,6 - In this case n is 5
22,23,24 - In this case n is 2.
Insufficient

Statement 2 - This gives us 2 values of n ie: 5 and 4. Insufficient

Combining - we can find m is a series of consecutive integers where n is 5. Just to contradict the case, you can check if you can get a series of 5 consecutive numbers where n is 4. I didn't find one.


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by GMATGuruNY » Thu Oct 04, 2012 8:00 am
das.ashmita wrote:What is the positive integer n?

(1) For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
(2) n^2 - 9n + 20 = 0

OA in the source is C

But I think its E
Statement 1: For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
Since m is positive, m(m+1)(m+2)...(m+n) represents the product of CONSECUTIVE positive integers.
Every other consecutive EVEN integer is a MULTIPLE OF 4.
Thus, if the product includes 3 consecutive EVEN factors, there are two cases:

Case 1: (multiple of 2)(multiple of 4)(multiple of 2)
This product must be divisible by (2)(4)(2) = 16.

Case 2: (multiple of 4)(multiple of 2)(multiple of 4)
This product must be divisible by (4)(2)(4) = 32.

In each case, the product is divisible by 16.
Thus, if the product includes at least 3 consecutive even factors, it will be divisible by 16.

To GUARANTEE that the product will include at least 3 consecutive even factors, the MINIMUM value of n is 5:
m(m+1)(m+2)(m+3)(m+4)(m+5) = the product of 6 consecutive integers.
This product will be composed of exactly 3 consecutive odd factors and exactly 3 consecutive even factors.

If n=4 -- implying a total of 5 consecutive factors -- then the product could be (odd)(even)(odd)(even)(odd).
In this case, the product will not include at least 3 consecutive even factors, with the result that it might not be divisible by 16.
To illustrate:
Neither 1*2*3*4*5 nor 17*18*19*20*21 is divisible by 16.

Since the product must include at least 3 consecutive even factors to GUARANTEE that it will be divisible by 16, n≥5.
No way to determine the exact value of n.
INSUFFICIENT.

Statement 2: n² - 9n + 20 = 0
(n-4)(n-5) = 0.
Since it's possible that n=4 or n=5, INSUFFICIENT.

Statements 1 and 2 combined:
Only n=5 satisfies both statements.
SUFFICIENT.

The correct answer is C.
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by Anurag@Gurome » Thu Oct 04, 2012 9:15 pm
das.ashmita wrote:What is the positive integer n?

(1) For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
(2) n^2 - 9n + 20 = 0

OA in the source is C

But I think its E
Statement 1: Implies the product of any (n + 1) consecutive positive integers is divisible by 16. Now the product of any 6 or more consecutive integers is always divisible by 16. Hence, (n + 1) ≥ 6 => n ≥ 5

Not sufficient

Statement 2: n² - 9n + 20 = 0
=> (n - 4)(n - 5) = 0
Hence, n = 4 or n = 5

Not sufficient

1 & 2 Together: As n ≥ 5, n must be equal to 5.

Sufficient

The correct answer is C.
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by das.ashmita » Fri Oct 05, 2012 1:16 am
Thankyou Anurag , Mitch and Anup for replying.

I got confused in statement 1

case 1 : let m=2
2*3*4*5*6 is divisible by 16
Here n = 6-2 = 4

case 2 : let consider m=1
1*2*3*4*5*6 is divisible by 16
here n = 6-1 = 5

Can u please tell me what is wrong in assuming case 1
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by GMATGuruNY » Fri Oct 05, 2012 4:15 am
das.ashmita wrote:Thankyou Anurag , Mitch and Anup for replying.

I got confused in statement 1

case 1 : let m=2
2*3*4*5*6 is divisible by 16
Here n = 6-2 = 4

case 2 : let consider m=1
1*2*3*4*5*6 is divisible by 16
here n = 6-1 = 5

Can u please tell me what is wrong in assuming case 1
You have cited ONE CASE in which n=4 yields a product divisible by 16.
But statement 1 requires that the product be divisible by 16 FOR EVERY positive integer m.
If m=1 and n=4, the resulting product is 1*2*3*4*5, which is not divisible by 16.
Thus, n=4 does NOT satisfy the condition that the product be divisible by 16 FOR EVERY positive integer m.
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by das.ashmita » Fri Oct 05, 2012 8:19 am
thanks Mitch!!absolutely clear now :)
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