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Unable to understand this DS problem

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Unable to understand this DS problem

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If P has a total of 8 positive factors, including 1 and P, what is the value of P? Positive integer
P has 2 positive prime factors, 5 and 11.

1. 125 is a factor of P.
2. 121 is not a factor of P.

I can understand that statement 1 is sufficient to answer the question because we can determine the 8 positive factors as follows :

statement 1: 125 can be written as 5^3 .So 5^2 ,5 ,11,11*5,11*5^2,11*5^3,1 are all factors of P.Since p is a factor of itself P=11*5^3.

But the answer says that statement 2 is also sufficient.How to determine that?

Thanks for your help.



Last edited by manjus_mailme on Wed Apr 14, 2010 12:22 pm; edited 1 time in total

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manjus_mailme wrote:
If P has a total of 8 positive factors,.....
P has exactly 2 positive prime factors, 5 and 11.

Please correct the question

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eaakbari wrote:
manjus_mailme wrote:
If P has a total of 8 positive factors,.....
P has exactly 2 positive prime factors, 5 and 11.

Please correct the question
Does the question make sense now.The question from the source itself was wrong.I edited it as it is now.

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Let's start by fixing up the problem:

Quote:
If P has a total of 8 positive factors, including 1 and P, what is the value of P? Positive integer
P has 2 positive prime factors, 5 and 11.
becomes:

Quote:
Positive integer P has only 2 distinct prime factors, 5 and 11. If P has a total of 8 positive factors, what's the value of P?
Step 1 of the Kaplan Method for DS: Analyze the stem

We know that 4 of the factors of P are 1, 5, 11 and P. We also know that P has no other primes factors other than 5 and 11; what we don't know is how many 5s and 11s P contains.

Step 2 of the Kaplan Method for DS: Evaluate the Statements

(1) 125 is a factor of P.

Therefore, P has at least 3 "5"s among it's primes.

So, we have 1, 5, 11, 25, 55, 125, 25*11, 125*11 as our 8 factors. Therefore, P = 125*11... sufficient.

(2) 121 is NOT a factor of P.

Therefore, P has exactly 1 factor of 11; the other factors all have to be 5s.

Well, the only way to generate 8 factors for P is if we have 3 "5"s among the factors; as above, P = 125*11... sufficient.

Each of (1) and (2) is sufficient alone, choose (D).

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P has 8 factors
1, 5 , 11 , a, b, c , d, P
(1) 125 is a factor of P.
so a, b, c or d can be 125
so possible values of a,b and c are
25, 55, 125, 25*11
( but does it have to be 55 or 25*11 can it not be 121 )
since there are 4 more factors a to d, does it restrict the existence of 11*11 as one of the factors
till (2) 121 is NOT a factor of P clears the air.
and if it does restrict the existence of 121 then is (2) really necessary.

What am I missing ?

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If a Number P is factorized in the format P= a^m * b^n.

and a , b are two prime factors of a number, then the number of factors is given by = (m+1)*(n+1).

For example, No of factors of 6 = 2^1*3^1 = (1+1)*(1+1)=2*2=4 ----> 1,2,3,6.

From the Question,

we have 5 and 11 as prime factors.... 5^m * 11^n ...

And Number of factors=8

From (i)

125 is factor of P---> 125=5^3 and P=5^3 * 11^n

Number of factors = (3+1)*(n+1)=8---> n=1.

S(i) is Sufficient.



From (ii)

121 is not a factor of P.

Only possibility=5^3*11---> Sufficient.

Hence D.

Thx Stuart.

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Thank you so much for explaning the problem .It was very helpful.

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