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# What is the positive integer n?

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What is the positive integer n? [#permalink]  27 Jan 2012, 15:08
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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

[Reveal] Spoiler:
Guys - the OA is C. But can someone please let me know what does statement 1 implies over here?

For me it says that the product of consecutive integers is divisible by 16. But how its used in this question?

I understand statement 2 though.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 01 Oct 2012, 04:47, edited 1 time in total.
Renamed the topic and edited the question.
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Re: +ve integer n [#permalink]  27 Jan 2012, 15:38
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enigma123 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

Guys - the OA is C. But can someone please let me know what does statement 1 implies over here?

For me it says that the product of consecutive integers is divisible by 16. But how its used in this question?

I understand statement 2 though.

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 --> as the given product is divisible by 16 for EVERY positive integer then the least value of n will be for the worst case scenario, which is for m=1, then m(m + 1)(m + 2) ... (m + n) to be divisible by 16 n must be at least 5: 1*2*3*4*5*6=divisible by 16 (as we have three even numbers out of which one is also a multiple of 4, so the product has 2*2^2*2=16 as a factor), but if n=4 and m=1 then 1*2*3*4*5 is not divisible by 16. Naturally if n is more than 5, then the product m(m + 1)(m + 2) ... (m + n) will also be divisible by 16 for the same reason: there will be at least 3 evens out of which one is also a multiple of 4. So from this statement we have that: $$n\geq{5}$$. Not sufficient.

Side note: we basically have the product of n+1 consecutive integers which is always divisible by (n+1)! and it to be divisible by 16, n must be more than or equal to 5.

(2) n^2 - 9n + 20 = 0 --> n=4 or n=5. Not Sufficient.

(1)+(2) From (1) $$n\geq{5}$$ and from (2) $$n=4$$ or $$n=5$$ --> $$n=5$$. Sufficient.

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Re: What is the positive integer n? (1) For every positive [#permalink]  01 Oct 2012, 02:41
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enigma123 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

Guys - the OA is C. But can someone please let me know what does statement 1 implies over here?

For me it says that the product of consecutive integers is divisible by 16. But how its used in this question?

I understand statement 2 though.

Usually, when approaching a DS question, you start with statement (1), then go to statement (2)...
If you don't know how to attack statement (1), go directly to (2), especially if it looks "friendlier", as in this case.

(2) The given quadratic equation has two solutions: $$n=4$$ and $$n=5$$.
Not sufficient. So, the answer is certainly not B.

Now go back to (1) and start by checking the two values for $$n$$ you obtained in (2).
Take $$m=1$$. Obviously, $$1\cdot2\cdot3\cdot4\cdot5$$ is not divisible by 16, but $$1\cdot2\cdot3\cdot4\cdot5\cdot6$$ is.
The answer can be C, if without the condition in statement (2), there can be more than one value of $$n$$ which fulfills the condition.
You don't have to look for the smallest $$n$$. Obviously, if $$n=15$$, we have $$16$$ consecutive numbers, so their product is definitely divisible by 16.
Therefore, the answer cannot be A.

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Re: What is the positive integer n? (1) For every positive [#permalink]  30 Sep 2012, 21:31
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enigma123 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

Guys - the OA is C. But can someone please let me know what does statement 1 implies over here?

For me it says that the product of consecutive integers is divisible by 16. But how its used in this question?

I understand statement 2 though.

Product of consecutive integers have some properties. e.g. product of any two consecutive positive integers is even, product of any 3 consecutive integers is even and is divisible by 3 so basically the product is divisible by 6. Why and how? Check out these posts:
http://www.veritasprep.com/blog/2011/09 ... c-or-math/
http://www.veritasprep.com/blog/2011/09 ... h-part-ii/

The first statement just tells you that every product of (n+1) consecutive integers is always divisible by 16.

Say if n = 3
Is 1*2*3*4 divisible by 16? No! So n cannot be 3.

Is 1*2*3....*14*15*16 divisible by 16? Yes. Will product of any 16 consecutive integers be divisible by 16? Yes. Product of any 17 consecutive integers will also be divisible by 16. So n can take many values.

What is the smallest value that n can take?
Every set of 4 consecutive integers will have a number which has 4 as a factor and it will have another even number i.e. the product of 4 consecutive numbers must be divisible by 8.
When you have product of 5 consecutive factors, again the product must be divisible by at least 8 e.g. 1*2*3*4*5
When you have 6 consecutive factors, there must be a number with 4 as a factor and 2 other even numbers i.e. the product must have 16 as a factor e.g. 1*2*3*4*5*6
Therefore, you must have at least 6 consecutive integers i.e. n must be at least 5.

@fameatop: n cannot be 4. If you have 5 consecutive factors, the product will not be divisible by 16 in every case e.g. it is not divisible in case of 1*2*3*4*5

Statement 1 tells you that n is at least 5
Statement 2 tells you that n is 4 or 5.

So n must be 5 using both
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Re: What is the positive integer n? [#permalink]  03 Oct 2012, 09:47
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ashd wrote:
Hi

From statement 1, if we consider m=2,
2*3*4*5*6 will be divisible by 16.
and here m+n=6 => n=4

So IMO the ans should be E

please correct me if I am wrong.

Statement (1) says: For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16.
If m = 1 and n = 4, 1*2*3*4*5 is not divisible by 16. So n cannot be 4.
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Re: What is the positive integer n? [#permalink]  03 Oct 2012, 19:20
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ashd wrote:
Quote:
Statement (1) says: For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16.
If m = 1 and n = 4, 1*2*3*4*5 is not divisible by 16. So n cannot be 4.

Hi

Again my question, why are we only considering the case where m=1?
why cant we consider m = 2 or the series starting from 2
In this case, 2*3*4*5*6 is divisible by 16

It is a technique we use. We want to find the value of n which holds for every possible value of m. When you want to prove that a particular value of n will hold for every value of m, it is very hard to do. You possibly cannot calculate for EVERY value of m. So we use the reverse logic. We assume that n takes a particular value and then try to find one value of m for which it doesn't hold. Then we can say that n cannot take that particular value.

Here, we are saying that if n = 4, it doesn't hold for m = 1. It doesn't matter whether it holds for other values of m or not. As long as there is even a single value of m for which it doesn't hold, we know n cannot be 4.

When n = 5, we see that it holds for quite a few values of m. Now the question is, will it hold for every value of m? For that, we use our theoretical knowledge to say that yes, it will hold for every value of m.
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Re: +ve integer n [#permalink]  30 Sep 2012, 14:52
Bunuel wrote:
enigma123 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

Guys - the OA is C. But can someone please let me know what does statement 1 implies over here?

For me it says that the product of consecutive integers is divisible by 16. But how its used in this question?

I understand statement 2 though.

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 --> as the given product is divisible by 16 for EVERY positive integer then the least value of n will be for the worst case scenario, which is for m=1, then m(m + 1)(m + 2) ... (m + n) to be divisible by 16 n must be at least 5: 1*2*3*4*5*6=divisible by 16 (as we have three even numbers out of which one is also a multiple of 4, so the product has 2*2^2*2=16 as a factor), but if n=4 and m=1 then 1*2*3*4*5 is not divisible by 16. Naturally if n is more than 5, then the product m(m + 1)(m + 2) ... (m + n) will also be divisible by 16 for the same reason: there will be at least 3 evens out of which one is also a multiple of 4. So from this statement we have that: $$n\geq{5}$$. Not sufficient.

Side note: we basically have the product of n+1 consecutive integers which is always divisible by (n+1)! and it to be divisible by 16, n must be more than or equal to 5.

(2) n^2 - 9n + 20 = 0 --> n=4 or n=5. Not Sufficient.

(1)+(2) From (1) $$n\geq{5}$$ and from (2) $$n=4$$ or $$n=5$$ --> $$n=5$$. Sufficient.

Hi bunuel,
I have a doubt in this question.
Value of m Minimum value of n
1 5
2 4
3 5
4 4
5 4
6 3
7 4
8 3
9 4
10 3
11 4
12 3
1) Thus n can be $$n<=5$$-->Insufficient
2) n can be 4 or 5 -Insufficient
1+2) n can take both 4 or 5 - Insufficient

So as per me answer has to be E, & if i am wrong where am i making mistake.
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Re: What is the positive integer n? [#permalink]  03 Oct 2012, 08:54
Hi

From statement 1, if we consider m=2,
2*3*4*5*6 will be divisible by 16.
and here m+n=6 => n=4

So IMO the ans should be E

please correct me if I am wrong.
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Re: What is the positive integer n? [#permalink]  03 Oct 2012, 10:46
Quote:
Statement (1) says: For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16.
If m = 1 and n = 4, 1*2*3*4*5 is not divisible by 16. So n cannot be 4.

Hi

Again my question, why are we only considering the case where m=1?
why cant we consider m = 2 or the series starting from 2
In this case, 2*3*4*5*6 is divisible by 16
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Re: What is the positive integer n? [#permalink]  03 Oct 2012, 11:18
ashd wrote:
Quote:
Statement (1) says: For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16.
If m = 1 and n = 4, 1*2*3*4*5 is not divisible by 16. So n cannot be 4.

Hi

Again my question, why are we only considering the case where m=1?
why cant we consider m = 2 or the series starting from 2
In this case, 2*3*4*5*6 is divisible by 16

We are not considering only the case m=1. For a given n, the product m(m+1)(m+2)...(m+n) must be divisible by 16 for m=1, also for m=2, for m=3, ..., for every integer m. But if for a specific value of m the product is not divisible by 16, that particular n is not an acceptable value. Since for m=1 the product is not divisible by 16 when n=4, we can conclude that n cannot be 4. The property doesn't for for m=9 either: 9*10*11*12*13. If you continue, you can find even more values for m for which the property will not hold when n=4.
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Re: What is the positive integer n? [#permalink]  14 Aug 2013, 01:24
Expert's post
Bumping for review and further discussion.
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Re: What is the positive integer n? (1) For every positive [#permalink]  11 Sep 2013, 23:03
EvaJager wrote:
enigma123 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

Guys - the OA is C. But can someone please let me know what does statement 1 implies over here?

For me it says that the product of consecutive integers is divisible by 16. But how its used in this question?

I understand statement 2 though.

Usually, when approaching a DS question, you start with statement (1), then go to statement (2)...
If you don't know how to attack statement (1), go directly to (2), especially if it looks "friendlier", as in this case.

(2) The given quadratic equation has two solutions: $$n=4$$ and $$n=5$$.
Not sufficient. So, the answer is certainly not B.

Now go back to (1) and start by checking the two values for $$n$$ you obtained in (2).
Take $$m=1$$. Obviously, $$1\cdot2\cdot3\cdot4\cdot5$$ is not divisible by 16, but $$1\cdot2\cdot3\cdot4\cdot5\cdot6$$ is.
The answer can be C, if without the condition in statement (2), there can be more than one value of $$n$$ which fulfills the condition.
You don't have to look for the smallest $$n$$. Obviously, if $$n=15$$, we have $$16$$ consecutive numbers, so their product is definitely divisible by 16.
Therefore, the answer cannot be A.

Found it really understandable.. Thank You So much Sir .

if (1) is not friendly go to (2) and take out possible values..it will save lot of time and thinking..

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Re: +ve integer n [#permalink]  15 Jun 2014, 18:00
fameatop wrote:
Bunuel wrote:
enigma123 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

Guys - the OA is C. But can someone please let me know what does statement 1 implies over here?

For me it says that the product of consecutive integers is divisible by 16. But how its used in this question?

I understand statement 2 though.

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 --> as the given product is divisible by 16 for EVERY positive integer then the least value of n will be for the worst case scenario, which is for m=1, then m(m + 1)(m + 2) ... (m + n) to be divisible by 16 n must be at least 5: 1*2*3*4*5*6=divisible by 16 (as we have three even numbers out of which one is also a multiple of 4, so the product has 2*2^2*2=16 as a factor), but if n=4 and m=1 then 1*2*3*4*5 is not divisible by 16. Naturally if n is more than 5, then the product m(m + 1)(m + 2) ... (m + n) will also be divisible by 16 for the same reason: there will be at least 3 evens out of which one is also a multiple of 4. So from this statement we have that: $$n\geq{5}$$. Not sufficient.

Side note: we basically have the product of n+1 consecutive integers which is always divisible by (n+1)! and it to be divisible by 16, n must be more than or equal to 5.

(2) n^2 - 9n + 20 = 0 --> n=4 or n=5. Not Sufficient.

(1)+(2) From (1) $$n\geq{5}$$ and from (2) $$n=4$$ or $$n=5$$ --> $$n=5$$. Sufficient.

Hi bunuel,
I have a doubt in this question.
Value of m Minimum value of n
1 5
2 4
3 5
4 4
5 4
6 3
7 4
8 3
9 4
10 3
11 4
12 3
1) Thus n can be $$n<=5$$-->Insufficient
2) n can be 4 or 5 -Insufficient
1+2) n can take both 4 or 5 - Insufficient

So as per me answer has to be E, & if i am wrong where am i making mistake.

Hi bunuel,
I have a doubt in this question.
Value of m Minimum value of n
1 5
2 4
3 5
4 4
5 4
6 3
7 4
8 3
9 4
10 3
11 4
12 3
1) Thus n can be n<=5-->Insufficient
2) n can be 4 or 5 -Insufficient
1+2) n can take both 4 or 5 - Insufficient

So as per me answer has to be E, & if i am wrong where am i making mistake.
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Re: What is the positive integer n? [#permalink]  15 Jun 2014, 21:40
enigma123 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

[Reveal] Spoiler:
Guys - the OA is C. But can someone please let me know what does statement 1 implies over here?

For me it says that the product of consecutive integers is divisible by 16. But how its used in this question?

I understand statement 2 though.

Statement I is insufficient:
If n = 8 then m(m+1)(m+2)...(m+8) will have at least 4 even integers which means the product will have to be divisible by 16
same goes for n = 9, n = 10 etc....

Statement II is insufficient:
n^2 - 5n - 4n + 20 = 0
n(n-5) - 4( n - 5) = 0
(n-5) (n-4) = 0

We get two value of n, hence not sufficient.

Combining is sufficient:

If we say that n = 4 then m (m+1) (m+2) (m+3) (m+4) will not be divisible by 16 if m = 1 (1 (2)(3)(4)(5))
If we say that n = 5 then m (m+1) (m+2)(m+3) (m+4)(m+5) will always be divisible by 16. Lets try m = 1(2)(3)(4)(5)(6) as there will be a minimum of 3 even numbers with a multiple of 4 between them. Hence a factor of 16 is a must.

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Re: What is the positive integer n?   [#permalink] 15 Jun 2014, 21:40
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