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What is the positive integer n? [#permalink]
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16 Aug 2007, 17:31
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What is the positive integer n? (1) For every integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16 (2) n^2  9n + 20 = 0 OPEN DISCUSSION OF THIS QUESTION IS HERE: whatisthepositiveintegern1foreverypositive126636.html
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Last edited by Bunuel on 02 Aug 2012, 13:11, edited 1 time in total.
OA added.



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I don't think I understand statement 1 but I would be forced to guess C.
Statement two tells us that N is either 4 or 5. and since statement 1 mentions 16 I would guess that the two together would somehow tell us it's 4!
haha, sorry I have no idea. just spitting out my logic



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St2:
n = 4 or 5. insufficient.
st1:
the product is divisible by 16. say the product is x, then we can write x/16 = 2/2^4. so as long as x has four 2's, we can cancel out 16. the statement says this works for every integer m, so let's say m=1. if m = 5, then x =720 which is divisible by 16. if m = 4, then x = 120, which is not divisible by 16. if m = 7, then x = 2520 which is also divisible by 16. Insufficient.
using st2 and st1, we know n is 5. Sufficient.
Ans C



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Re: DS: How Many m and n's? [#permalink]
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16 Aug 2007, 21:07
sludge wrote: What is the positive integer n?
(1) For every integer m, the product m (m + 1) (m + 2) ... (m + n) is divisible by 16
(2) n^2  9n + 20 = 0
got E.
from 1: m can be any integer but n varies depending upon m till the expression [m(m + 1)(m + 2) ... (m + n)] is divisible by 16.
so if m = 1, n = 5.
if m = 2, n = 4
if m = 3, n = 5
if m = 4, n = 5
if m = 5, n = 4
so, m could be 4 or 5.
from 2: n = 4 or 5.
from 1 and 2 also n = 4 or 5.
so annswer should be E



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Got E.
I used the same method as Fistail:
If m=1, then n > 5
If m=2, then n > 4
and so on...
So (1) is insufficient.
For (2), n = 4 or 5. So (2) is insufficient as well.
(1) and (2) doesn't provide if n is 4 or 5 so E.
Is there a way to do this problem faster?



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Re: DS: How Many m and n's? [#permalink]
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17 Aug 2007, 02:49
sludge wrote: What is the positive integer n?
(1) For every integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
(2) n^2  9n + 20 = 0
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. (B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not. (C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient to answer the question. (E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
(1) tells us that n is at least 5, as the product of 6 or more consecutive integers is always a multiple of 16. The product of fewer than 6 consecutive integers need not be a multiple of 16.
NOT SUFF
(2) n is either 4 or 5
NOT SUFF
(T) n=5
SUFF



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Re: DS: How Many m and n's? [#permalink]
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17 Aug 2007, 07:03
kevincan wrote: sludge wrote: What is the positive integer n?
(1) For every integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
(2) n^2  9n + 20 = 0
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. (B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not. (C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient to answer the question. (E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question. (1) tells us that n is at least 5, as the product of 6 or more consecutive integers is always a multiple of 16. The product of fewer than 6 consecutive integers need not be a multiple of 16. NOT SUFF (2) n is either 4 or 5 NOT SUFF (T) n=5 SUFF
if m=2, m(m+1)(m+2)(m+3)(m+4) =(2x3x4x5x6) is divided by 16.
so n here is 4.



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However, is m(m+1)(m+2)(m+3)(m+4) a multiple of 16 for every integer m?



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kevincan wrote: However, is m(m+1)(m+2)(m+3)(m+4) a multiple of 16 for every integer m?
nope. in this case n = 4, in some other cases n=5 or could be more or less.
if m = 1, n = 5.
if m = 8, n = 3
if m = 15, n =2
if m = 16, n =1.
the divisibility of the expression m(m+1)(m+2)(m+3)(m+4) by depends upon n.
therefore it is E.
imo, this is really good question.



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Re: DS: How Many m and n's? [#permalink]
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17 Aug 2007, 09:04
sludge wrote: What is the positive integer n?
(1) For every integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
(2) n^2  9n + 20 = 0
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. (B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not. (C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient to answer the question. (E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
I think it's 'A'. n=15 always holds good, it's not really asking for a minimum value of n.



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Re: DS: How Many m and n's? [#permalink]
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17 Aug 2007, 10:38
sludge wrote: What is the positive integer n?
(1) For every integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
(2) n^2  9n + 20 = 0
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. (B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not. (C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient to answer the question. (E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
C for me.
16= 2*2*2*2
(1) If we know that there are at least 4 even numbers, then you know that it will be divisible by 16. However, you don't have to have 4 even numbers because if m=2, then next even term is m=4. The next even term after that is m=6. This will only require 3 even terms to satisfy the equation; thus, if m is odd, n=6 will satisfy the equation. If m is even, n=4 will satisfy the equation. In sum, n=4 or n=6
INSUFFICIENT.
(2) n=5 or n=4. INSUFFICIENT.
Together, we know that n=4
SUFFICIENT



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Fistail wrote: dahcrap wrote: It is clearly C how is that? i wait for OA.
From 1st alone we cant say. n can has so many values depends on m.
From 2nd option we got n 4 and 5...
2nd is also insufficient...
When we put both value of n in 1st.It satisfied quation on diff value of m not every value of m.
We are trying only +ve value of m.. m can be ve also.When we put m as 1,2,3,4 and n as 4 or 5.we get the product m(m + 1)(m + 2) ... (m + n) =0 which is not divisible by 16.
Answer should be E....



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Here's my attempt at it:
1. Quote: For every integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
this doesn't tell you anything specific about n. plug in 1 for m (since 1 is simple and M can be any integer) and see what works. 1(2) = 2 1(2)(3) = 6 1(2)(3)(4) = 24 1(2)(3)(4)(5) = 120 1(2)(3)(4)(5)(6) = 720 720 is divisible by 16, which means n = 5 works. this is just one example of an N that works, so you can't be sure that 5 is it, but you can be sure that 1, 2, 3 and 4 do not work (since 2, 6, 24, 120 are not divisible by 16) as you can see, this is the same as (n+1)! = 16n. We know that 5 works, but there are tons of other possibilities greater than 5 that could be divisible by 16. Since it's DS and not PS we don't need to find them, just know that they exist and that A is NOT SUFFICIENT. 2. Quote: (2) n^2  9n + 20 = 0
break this down:
(n4)(n5) = 0
this means that N is = to either 4 or 5. obviously this isn't sufficient on it's own since we get two possible answers. NOT SUFFICIENT
but put them together and see what happens!
we already know that 4 does not work from the first statement, but 5 does. this gives 5 as the only possible answer and we get C is SUFFICIENT.
now if you went through statement 1 without testing for possibilities you could see from statement 2 that the answer is 4 or 5. then just go back to statement one and plug in 4 and 5 and see if one of them works.
m = 1 n = 4
1(1+1)(1+2)(1+3)(1+4)
1(2)(3)(4)(5) = 120
120 is not divisible by 16.
m = 1 n = 5
1(1+1)(1+2)(1+3)(1+4)(1+5)
1(2)(3)(4)(5)(6) = 720
720/16 = 46
so using both statements we see that the only possible answer is 5. C is sufficient!



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Re: DS: How Many m and n's? [#permalink]
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17 Aug 2007, 10:47
kevincan wrote: sludge wrote: What is the positive integer n?
(1) For every integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
(2) n^2  9n + 20 = 0
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not. (B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not. (C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient to answer the question. (E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question. (1) tells us that n is at least 5, as the product of 6 or more consecutive integers is always a multiple of 16. The product of fewer than 6 consecutive integers need not be a multiple of 16. NOT SUFF (2) n is either 4 or 5 NOT SUFF (T) n=5 SUFF
I got the same answer, but n can be less than 5 for (1). For example, say m=2 and n=4. You have: 2*3*4*5*6, which satisfy the multiplication.
In fact, if m is even, you only need n=4. This is true for all cases. Try it out.



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m negative [#permalink]
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17 Aug 2007, 11:00
Why are you guys not considering the vaule of m negative....
M is juss a interger.It can be negative and zero also
The integers (Latin, integer, literally, "untouched," whole, entire, i.e. a whole number) are the positive natural numbers (1, 2, 3, …), their negatives (−1, −2, −3, ...) and the number zero.



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Re: m negative [#permalink]
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17 Aug 2007, 11:01
chiya wrote: Why are you guys not considering the vaule of m negative....
M is juss a interger.It can be negative and zero also
The integers (Latin, integer, literally, "untouched," whole, entire, i.e. a whole number) are the positive natural numbers (1, 2, 3, …), their negatives (−1, −2, −3, ...) and the number zero.
I consider m negative and zero. I break it down to even and odd, which consider all.



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Re: m negative [#permalink]
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17 Aug 2007, 11:07
chiya wrote: Why are you guys not considering the vaule of m negative....
M is juss a interger.It can be negative and zero also
The integers (Latin, integer, literally, "untouched," whole, entire, i.e. a whole number) are the positive natural numbers (1, 2, 3, …), their negatives (−1, −2, −3, ...) and the number zero.
because the stem tells us that N is positive.
and Statement 1 tells us that "for every integer m, the product..."
so we're looking for a positive n and we can plug any integer we want into the equation. plugging in 1, 2, 3 etc will just give us 0 for the answer as long as n > m. 1(1 +1) = 1(0), and so on
and since 0 is divisible by everything it will work for 16



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Re: m negative [#permalink]
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17 Aug 2007, 11:07
bkk145 wrote: chiya wrote: Why are you guys not considering the vaule of m negative....
M is juss a interger.It can be negative and zero also
The integers (Latin, integer, literally, "untouched," whole, entire, i.e. a whole number) are the positive natural numbers (1, 2, 3, …), their negatives (−1, −2, −3, ...) and the number zero. I consider m negative and zero. I break it down to even and odd, which consider all.
But when m = 0 the product m(m + 1)(m + 2) ... (m + n) =0 which is not divisible by 16.



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I agree with Fistail here and my pick is E.
According to 1st statement M can be any integer thus there is no fiexed value for n.
from 2. n can be 4 or 5.
Thus it has to be E. Unless i am missing some logic in statement 1 but i cannot see any way where we can assign a fixed value for m or n.
I think this is one of 700+ questions where you tend to pick C because you know it is a high score question.



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excelgmat wrote: I agree with Fistail here and my pick is E. According to 1st statement M can be any integer thus there is no fiexed value for n. from 2. n can be 4 or 5.
Thus it has to be E. Unless i am missing some logic in statement 1 but i cannot see any way where we can assign a fixed value for m or n. I think this is one of 700+ questions where you tend to pick C because you know it is a high score question.
(1) For every integer m, m(m+1)...(m+n) is a multiple of 16 for EVERY integer m if and only if n>4. (1) tells us that n is at least 5. NOT SUFF
(2) n is either 4 or 5 NOT SUFF
(T) n=5 SUFF







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