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What is positive integer n?

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Source: — Data Sufficiency |

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by Anurag@Gurome » Fri Mar 18, 2011 10:27 am
RadiumBall wrote:What is 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.
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 Night reader » Fri Mar 18, 2011 11:14 am
n (n is integer) >0, n-?
st(1) m (m is integer) >0; m!/i =16, where (i is integer); m can be any value, hence n can be any value --> 1*2*3*4*5*6/45=16 [6!/45 =16], m=1, n=5, m(m+1)(m+...5) OR 7!/315=16, m=1, n=6 Not Sufficient;

st(2) By solving the quadratic equation we need to find Only One value for x; if we have found two values, statement(2) is Not Sufficient. Let's check --> D (discriminant) =sqrt(81-80), n(1,2)=(9+-1)/2=(4;5) we found two values-hence Not Sufficient;

Combined st(1&2): we have two options n=4 and n=5 out of these two options we can plug into m(m+1)(m+...n) and check for n ---> n={4;5} m(m+1)(m+2)(m+3)(m+n)=16*i. On the right side we have 16 and i where i can be anything Not Sufficient

IOM E

check point for option E ---> 16*9!=m(m+1)(m+2)(m+3)(m+...n) where m=5 and n=4
5(5+1)(5+2)(5+3)(5+4)=5*6*7*8*9; 5(5+1)(5+2)(5+3)(5+4)/16 = 945 OR m=4 and n=5, 4(4+1)(4+2)(4+3)(4+4)(4+5)=4*5*6*7*8*9; 4(4+1)(4+2)(4+3)(4+4)(4+5)/16=3,780

n is said to be both n=4 and n=5 with condition of st(1) m(m+1)(m+2)(m+3)(m+...n)/16 where n and m are positive integers;

I decided on E
RadiumBall wrote:What is 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.
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by MAAJ » Fri Mar 18, 2011 2:40 pm
Anurag@Gurome wrote:
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.
Can you elaborate Statement one a little bit further?

I guessed (C) but I'm pretty confused about how to solve this one.... I arrived to a dead end but still want to know if I was on the right/wrong way to solve this.... Here's what I did:

m(m+1)(m+2)...(m+n) is divisible by 16, so it has at least 4*4 in its factor primes

If N=4
m(m+1)(m+2)(m+3) is divisible by our first 4 because they are four consecutive numbers. That leaves us with (m+4)/4?, which translates to is m/4? ... Could be if m = 4,8,16...

If N=5
m(m+1)(m+2)(m+3) is divisible by our first 4, which leaves us to (m+4)(m+5)/4? Could be if m = 0,3,....

So what's the OA..?
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by Night reader » Fri Mar 18, 2011 4:20 pm
MAAJ wrote:...
m(m+1)(m+2)...(m+n) is divisible by 16, so it has at least 4*4 in its factor primes
why limit to 4*4?
MAAJ wrote:...
If N=5
m(m+1)(m+2)(m+3) is divisible by our first 4, which leaves us to (m+4)(m+5)/4? Could be if m = 0,3,....
m cannot be 0, as m>0
__________________________________________________________
Look, m(m+1)(m+2)(m+3)(m+...n) where m=5 and n=4

5(5+1)(5+2)(5+3)(5+4)/16 = 945

OR, m=4 and n=5

4(4+1)(4+2)(4+3)(4+4)(4+5)/16=3,780

As you see n still can be equal to two values.

Do you agree about answer E?
What is 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.
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by atulmangal » Fri Mar 18, 2011 9:15 pm
I agree with NR

If we combine both the choices, from Op B, we get a choice to select any value 4 or 5 for n
but for "m" even after combining we get nothing...that means m can be anything from 0 to infinity

So, for n=5, and m=1 (as m can be any positive int)

1*2*3*4*5*6 = 720 divisible by 16

but for n=4, and m=1
1*2*3*4*5 = 120 NOT Divisible

here it seems only n=5 is possible but

for n=4 and m=15 (again m can be any positive int, its a free bird)

15*16*17*18*19 = clearly divisible by 16....hence E
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by MAAJ » Fri Mar 18, 2011 9:42 pm
Night reader wrote:
MAAJ wrote:...
m(m+1)(m+2)...(m+n) is divisible by 16, so it has at least 4*4 in its factor primes
why limit to 4*4? It could be divisible by 4*4*x*y*z...etc but at least it has one pair of fours.

MAAJ wrote:...
If N=5
m(m+1)(m+2)(m+3) is divisible by our first 4, which leaves us to (m+4)(m+5)/4? Could be if m = 0,3,....
m cannot be 0, as m>0 Yeah, your right! thanks
Do you think my approach to solve this problem was fine?
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by Anurag@Gurome » Sat Mar 19, 2011 12:52 am
MAAJ wrote:Can you elaborate Statement one a little bit further?
First of all note that the phrase "For every positive integer m, the product m(m +1)(m + 2)...(m + n)" simply implies product of any (n + 1) consecutive positive integers.

Now, the product of any (n + 1) consecutive positive integers is always divisible by (n + 1)!.

So if we have to make the product of any (n + 1) consecutive positive integers divisible by 16, we have to choose n in such a way that 16 is a factor of (n + 1)!. Which is only possible if (n + 1) ≥ 6, i.e. n ≥ 5
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by Night reader » Sat Mar 19, 2011 8:48 am
I have to disagree, "For every positive integer m, the product m(m +1)(m + 2)...(m + n)" simply implies product of any (n + 1) consecutive positive integers might imply (n+1)! only if we had not a factor m in front of (n+1)!, BUT here's a factor m which breaks through the usual (n+1)! formula.
Anurag@Gurome wrote:
MAAJ wrote:Can you elaborate Statement one a little bit further?
First of all note that the phrase "For every positive integer m, the product m(m +1)(m + 2)...(m + n)" simply implies product of any (n + 1) consecutive positive integers.

Now, the product of any (n + 1) consecutive positive integers is always divisible by (n + 1)!.

So if we have to make the product of any (n + 1) consecutive positive integers divisible by 16, we have to choose n in such a way that 16 is a factor of (n + 1)!. Which is only possible if (n + 1) ≥ 6, i.e. n ≥ 5
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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by RadiumBall » Sun Mar 20, 2011 4:28 am
Sorry got busy. OA: C

But Anurag could you please explain why n >= 5, why not 4?

Thank you.
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by Anurag@Gurome » Sun Mar 20, 2011 4:46 am
RadiumBall wrote:But Anurag could you please explain why n >= 5, why not 4?
I have tried to explain that algebraically and methodically, now let me explain it by picking numbers.

For any value of m other than 1, the product m(m +1)(m + 2)...(m + n) is always divisible by 16 for both n = 4 and n = 5.

But if m = 1, then,
For n = 4 --> m(m +1)(m + 2)...(m + n) = 1*2*3*4*5 = 120 --> NOT divisible by 16
For n = 5 --> m(m +1)(m + 2)...(m + n) = 1*2*3*4*5*6 = 720 --> DIVISIBLE by 16
For n > 5 --> m(m +1)(m + 2)...(m + n) = 1*2*3*4*5*6*n = 720*n --> DIVISIBLE by 16

Hence for every positive integer m, the product m(m +1)(m + 2)...(m + n) is NOT divisible by 16 if n = 4. Hence, 4 cannot be a value of n.

But for every positive integer m, the product m(m +1)(m + 2)...(m + n) is DIVISIBLE by 16 if n ≥ 5. Hence, any integer greater than 5 can be a value of n.

But from statement 2, only two values of n are possible 4 and 5, out of which 4 is discarded as it doesn't satisfy the statement 1. Hence, n = 5.

Hope that helps.
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by MAAJ » Sun Mar 20, 2011 7:58 am
Now I get it... but god, it took me a hell lot of time to understand this. This problem should be on the 700~ right? I'm still trying to figure out the shortcut given by Anurag :(

m(m+1)(m+2)(m+3)(m+4)

1) O E O E O -> M = ODD. At least one even number should be divisible by 4 and the other by 2. So we have 2*2*2 = 8, which is not sufficient to say that it's divisible by 16. It could only be divisible by 16 if one even number is divisible by 4 two times and that happens when we arrive to 8 or a multiple of 8, and the lowest possible odd value for "m" in which we get a factor of 8 is 5 (5+3 = 8) THUS for all odd numbers M>=5 the equation is divisible by 16.

2 ) E O E O E -> M = EVEN. At least two even number should be divisible by 2, and the other should be divisible by 4. So we have 2*2*2*2 = 16. Thus, the equation must be divisible by 16 if m = even number.

Combining 1) and 2) M should be even or an odd number greater or equal to 5

m(m+1)(m+2)(m+3)(m+4)(m+5)

3) O E O E O E -> M = ODD. We have 3 even numbers so at least 2 of them should be divisible by 2 and the other by 4, thus this is divisible by 16.

4) E O E O E O -> M = EVEN. Same as above, divisible by 16.

Combining 3) and 4) M can be any number, and THUS this is the correct SCENARIO.

So "N" MUST BE 5!!! For m(m+1)(m+2)...(m+n) be ALWAYS divisible by 16 [spoiler]Correct Answer (C)[/spoiler]
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