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GPREP FACTORS DS

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GPREP FACTORS DS

by cramya » Sun Oct 26, 2008 8:21 am
The positive integer k has exactly 2 positive prime factors 3 and 7. If k has a total of 6 positive factors inlcuding 1 and k what is the value of k?

1) 3 ^ 2 is a factor of k

2) 7 ^ 2 is not a factor of k


[spoiler] OA: D [/spoiler]
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Source: — Data Sufficiency |
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by cramya » Sun Oct 26, 2008 8:24 am
Please explain the approach in detail and not just SUFF / INSUFF.
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Re: GPREP FACTORS DS

by yezz » Sun Oct 26, 2008 8:32 am
cramya wrote:The positive integer k has exactly 2 positive prime factors 3 and 7. If k has a total of 6 positive factors inlcuding 1 and k what is the value of k?

1) 3 ^ 2 is a factor of k

2) 7 ^ 2 is not a factor of k

FROM STEM

K = 3^A*7^B

FROM 1

INSUFF

A = 2 OR 3 OR ...

FROM 2

B = 1

BOTH

iNSUFF

E
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by cramya » Sun Oct 26, 2008 8:34 am
Yezz, thanks!

The OA is different(not putting what the OA is in this post(inside the psoiler in my original post) since I dont want to spoil the fun for others :D )
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by parallel_chase » Sun Oct 26, 2008 8:44 am
total number of factors of k = 6

prime factors of k = 3,7

for total to be 6 factors there can be two cases

Case I

3^2 * 7 , no. of factors = (2+1)*(1+1) = 6

Case II

3* 7^2, no. of factors = (1+1) * (2+1) = 6

Statement I
3 ^ 2 is a factor of k, therefore, k has to be 3^2 * 7 = 63
Sufficient.


Statement II
7 ^ 2 is not a factor of k, therefore, k has to be 3^2 * 7 = 63.
Sufficient.

Hence D.

Let me know if you still have any doubts.
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by cramya » Sun Oct 26, 2008 8:48 am
Missed the 21 (3*7) and I was thinking it could be somethign else other than 63 also.

Appreciate it!
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by uttara » Mon Oct 27, 2008 11:32 am
Hi parallel_chase
can u explain these lines
Case I

3^2 * 7 , no. of factors = (2+1)*(1+1) = 6

Case II

3* 7^2, no. of factors = (1+1) * (2+1) = 6
how did u get no. of factors = (2+1)*(1+1) = 6 ??
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by parallel_chase » Mon Oct 27, 2008 11:39 am
uttara wrote:Hi parallel_chase
can u explain these lines
Case I

3^2 * 7 , no. of factors = (2+1)*(1+1) = 6

Case II

3* 7^2, no. of factors = (1+1) * (2+1) = 6
how did u get no. of factors = (2+1)*(1+1) = 6 ??
Here is an example of how you can find no. of factors for any number.

36 has 9 factors

Firstly prime factorize
36 = 2*2*3*3 = 2^2 * 3^2
Now only look at the powers of the prime factors.
(2+1)*(2+1) = 3*3 = 9


Similarly, if you need to find the factors of 63
63 = 3*3*7 = 3^2 * 7^1

Now again look at the powers of 3 and 7
Add 1 to both and multiply

(2+1)*(1+1) = 3*2 = 6.

Hope this helps.
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by uttara » Mon Oct 27, 2008 11:58 am
Thanks. I was not aware of this formula....
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