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

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.

Let's start by fixing up the problem:

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:

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).

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 ?

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.

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