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# PS question

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Senior Manager
Joined: 29 Aug 2005
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03 Jun 2008, 22:48
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How do i solve these type of questions

can u please explain me the technique
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The world is continuous, but the mind is discrete

Senior Manager
Joined: 29 Aug 2005
Posts: 266

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03 Jun 2008, 23:15
alpha_plus_gamma wrote:
vdhawan1 wrote:
How do i solve these type of questions

can u please explain me the technique

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The world is continuous, but the mind is discrete

Manager
Joined: 03 Jun 2008
Posts: 131
Schools: ISB, Tuck, Michigan (Ross), Darden, MBS

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03 Jun 2008, 23:40
vdhawan1 wrote:
How do i solve these type of questions

can u please explain me the technique

Ans. For such questions plz follow the following technique.

The number of factors for a number say X is always equal to (a+1)(b+1)(c+1)..., where a,b,c,.. denote the power of the prime numbers that make this number.

Eg. 50 = 5 (Squared) * 2, here the prime numbers are 5 and 2, and their powers are 2 and 1 respectively. So a and b in this case are 2 and 1.

No of Factors of 50 = (2+1)(1+1) = 6

Similarly for ur question N, has 4 prime numbers as its factors.

Therefore number of Factors for N = (a+1)(b+1)(c+1)(d+1), given the options, a,b,c,d cannot take any other values except 1. SO the ans is 16.

Let me know if this is still not clear.
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Manager
Joined: 28 May 2008
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03 Jun 2008, 23:43
1
KUDOS
vdhawan1 wrote:
How do i solve these type of questions

can u please explain me the technique

It is 16.

I ll show you by an example
lets say the number = 2*3*5*7
we need to find factors.
clearly 1,2,3,5,7 are the factors. total number=5
now selecting the product 2 numbers at a time 4C2= 6
now selecting the product of 3 nos =4c3 =4
now the number n itself =1
sum it up = 16

There is a shorter methid
If n= a^m * b^n * c^p and so on where a, b, c are the prime factors
then total no of divisors= (m+1)*(n+1)*(p+1)

for eg here 2^1 * 3^1 * 5^1 *7^1
Total no of divisore = (1+1)(1+1)(1+1)(1+1)= 16

cheers !!
Manager
Joined: 31 Oct 2007
Posts: 53

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03 Jun 2008, 23:57
I thought 1 wasnt a prime number ??

Thanks for the great explanation on factorization though.
Manager
Joined: 28 May 2008
Posts: 93

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04 Jun 2008, 00:00
snoor wrote:
I thought 1 wasnt a prime number ??

Thanks for the great explanation on factorization though.

Hey snoor
1 is STILL not a prime number.
but 1 sure is a factor of any number, right?
lets say n=2
It has 2 factors , 1 and itself
Senior Manager
Joined: 29 Aug 2005
Posts: 266

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04 Jun 2008, 00:15
zeenie wrote:
vdhawan1 wrote:
How do i solve these type of questions

can u please explain me the technique

It is 16.

I ll show you by an example
lets say the number = 2*3*5*7
we need to find factors.
clearly 1,2,3,5,7 are the factors. total number=5
now selecting the product 2 numbers at a time 4C2= 6
now selecting the product of 3 nos =4c3 =4
now the number n itself =1
sum it up = 16

There is a shorter methid
If n= a^m * b^n * c^p and so on where a, b, c are the prime factors
then total no of divisors= (m+1)*(n+1)*(p+1)

for eg here 2^1 * 3^1 * 5^1 *7^1
Total no of divisore = (1+1)(1+1)(1+1)(1+1)= 16

cheers !!

thanks for the great explanation
its very clear now
_________________

The world is continuous, but the mind is discrete

Senior Manager
Joined: 29 Aug 2005
Posts: 266

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04 Jun 2008, 00:17
GMBA85 wrote:
vdhawan1 wrote:
How do i solve these type of questions

can u please explain me the technique

Ans. For such questions plz follow the following technique.

The number of factors for a number say X is always equal to (a+1)(b+1)(c+1)..., where a,b,c,.. denote the power of the prime numbers that make this number.

Eg. 50 = 5 (Squared) * 2, here the prime numbers are 5 and 2, and their powers are 2 and 1 respectively. So a and b in this case are 2 and 1.

No of Factors of 50 = (2+1)(1+1) = 6

Similarly for ur question N, has 4 prime numbers as its factors.

Therefore number of Factors for N = (a+1)(b+1)(c+1)(d+1), given the options, a,b,c,d cannot take any other values except 1. SO the ans is 16.

Let me know if this is still not clear.

Thanks for the great explanation
its clear now
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Joined: 04 May 2006
Posts: 1830
Schools: CBS, Kellogg

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04 Jun 2008, 00:29
zeenie wrote:
snoor wrote:
I thought 1 wasnt a prime number ??

Thanks for the great explanation on factorization though.

Hey snoor
1 is STILL not a prime number.
but 1 sure is a factor of any number, right?
lets say n=2
It has 2 factors , 1 and itself

I think, it should be confused. Let see the original:

If positive number n is product of 4 different prime numbers, including 1 and n,.

I think,
a. adding 1 and n to the total, so the total is 4 numbers.
b. if the above statement is correct, 1 and n are not necessaryly prime number.

what do you think?
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Joined: 28 May 2008
Posts: 93

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04 Jun 2008, 02:01
sondenso wrote:
zeenie wrote:
snoor wrote:
I thought 1 wasnt a prime number ??

Thanks for the great explanation on factorization though.

Hey snoor
1 is STILL not a prime number.
but 1 sure is a factor of any number, right?
lets say n=2
It has 2 factors , 1 and itself

I think, it should be confused. Let see the original:

If positive number n is product of 4 different prime numbers, including 1 and n,.

I think,
a. adding 1 and n to the total, so the total is 4 numbers.
b. if the above statement is correct, 1 and n are not necessaryly prime number.

what do you think?

Hey there..
you might have to reinterpret that bit of the question stem that says "positive number n is product of 4 different prime numbers, including 1 and n,
After all, a number that is a product of 4 prime numbers cannot be a prime number ! right?
Question not framed well I guess. The factors would include 1 and n.
CEO
Joined: 29 Mar 2007
Posts: 2523

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04 Jun 2008, 06:37
vdhawan1 wrote:
How do i solve these type of questions

can u please explain me the technique

Here is how I solved.

XYZW
XYZW

Draw lines from X to the others, repeat for Y, but don't draw to its own and to X (B/c already done).

We get 3+2+1 from this.

Now we also have 4 from X, Y,Z,W alone

So we have 3+2+1+4 now. Don't forget 1 and XYWZ.

Thats --> 3+2+1+4+1+1.

Now the number of groups of 3 is simply 4!/3! --> 4

3+2+1+4+1+1+4 = 16.
Manager
Joined: 28 Apr 2008
Posts: 110

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14 Nov 2008, 02:18
16

4c1+4c2+4c3+2=16

2 is for n and 1.

4c1 is for each of the prime pactors

4c2 is for any multiple of 2 of the four prime numbers

4c3 is for any multiple of 3 of the prime numbers

hope that helps
Manager
Joined: 14 Oct 2008
Posts: 160

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14 Nov 2008, 04:23
I agree with Sondenso, this question is not correct. GMAT would never include such ambiguous questions in exam.

Although the method everyone is trying to show here is correct (a+1)(b+1) ... but it doesn't apply to this qs as it stands. Whats the source of this qs vwdhawan1 ? Can you also post the explanation they give as the answer ?

Thanks.
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Joined: 29 Aug 2007
Posts: 2457

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14 Nov 2008, 14:15
sondenso wrote:
zeenie wrote:
snoor wrote:
I thought 1 wasnt a prime number ??

Thanks for the great explanation on factorization though.

Hey snoor
1 is STILL not a prime number.
but 1 sure is a factor of any number, right?
lets say n=2
It has 2 factors , 1 and itself

I think, it should be confused. Let see the original:

If positive number n is product of 4 different prime numbers, including 1 and n,.

I think,
a. adding 1 and n to the total, so the total is 4 numbers.
b. if the above statement is correct, 1 and n are not necessaryly prime number.

what do you think?

THE QUESTION SHOULD READ AS UNDER: If positive integer n is the product of 4 different prime numbers, how many factors does n have, including 1 and n,?

two ways to solve it:

1: 4c1+4c2+4c3+4c4 = 16
2: (1+1)(1+1)(1+1)(1+1) = 16
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Manager
Joined: 23 Jul 2008
Posts: 189

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14 Nov 2008, 17:04
then in that case don t you think the answer shud be 18
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14 Nov 2008, 17:23
hibloom wrote:
then in that case don t you think the answer shud be 18

Those are already included in 16.
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15 Nov 2008, 08:51
should have read carefully it is included
Re: PS question   [#permalink] 15 Nov 2008, 08:51
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