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Numbers Prop

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by heshamelaziry » Sun Nov 15, 2009 12:51 am
Positive integer P has exactly 2 positive prime factors, 5 and 11. If P has a total of 8 positive factors, including 1 and P, what is the value of P?

A) 125 is a factor of P.
B) 121 is not a factor of P.
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Source: — Data Sufficiency |

Re: Numbers Prop

by Brent@GMATPrepNow » Sun Nov 15, 2009 9:56 am
heshamelaziry wrote:Positive integer P has exactly 2 positive prime factors, 5 and 11. If P has a total of 8 positive factors, including 1 and P, what is the value of P?
A) 125 is a factor of P.
B) 121 is not a factor of P.
Great question.
First, there's a little rule you should know:
If the prime factorization of a number, n, is such that n = p^a * q^b * r^c etc where p, q, r (etc) are prime numbers, then the number of positive divisors (factors) that n has is (a+1)(b+1)(c+1) etc
(e.g., 600=2^3 * 3^1 * 5*2, so 600 has 24 positive divisors since(3+1)(1+1)(2+1)=24)

We are told that P has two unique prime factors (5 and 11), which means that we can write P as follows: P = 5^x * 11^y, where x and y are both positive integers.
Our goal here is to find the value of P. In other words, we need to find the values of x and y.

Since we are told that P has exactly 8 positive divisors, we know (from the above rule) that (x+1)(y+1)=8
Since x and y must be positive integers, there are only two possible sets of values that satisfy the equation (x+1)(y+1)=8. The two possible solutions are i) x=1 and y=3, and ii) x=3 and y=1.

(A) if 125 (aka 5^3) is a factor of P, then we can conclude that x is 3 or greater. This means that x=3 and y=1, since we can now rule out the solution x=1 and y=3. SUFFICIENT

The answer is D

(B) If 121 (aka 11^2) is not a divisor of P, we know that y must be less than 2. In other words y must equal 1, in which case our only possible solution is x=3 and y=1. SUFFICIENT
Brent Hanneson - Creator of GMATPrepNow.com
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by mehravikas » Sun Nov 15, 2009 6:56 pm
Superb explanation...!!
Brent Hanneson wrote:
heshamelaziry wrote:Positive integer P has exactly 2 positive prime factors, 5 and 11. If P has a total of 8 positive factors, including 1 and P, what is the value of P?
A) 125 is a factor of P.
B) 121 is not a factor of P.
Great question.
First, there's a little rule you should know:
If the prime factorization of a number, n, is such that n = p^a * q^b * r^c etc where p, q, r (etc) are prime numbers, then the number of positive divisors (factors) that n has is (a+1)(b+1)(c+1) etc
(e.g., 600=2^3 * 3^1 * 5*2, so 600 has 24 positive divisors since(3+1)(1+1)(2+1)=24)

We are told that P has two unique prime factors (5 and 11), which means that we can write P as follows: P = 5^x * 11^y, where x and y are both positive integers.
Our goal here is to find the value of P. In other words, we need to find the values of x and y.

Since we are told that P has exactly 8 positive divisors, we know (from the above rule) that (x+1)(y+1)=8
Since x and y must be positive integers, there are only two possible sets of values that satisfy the equation (x+1)(y+1)=8. The two possible solutions are i) x=1 and y=3, and ii) x=3 and y=1.

(A) if 125 (aka 5^3) is a factor of P, then we can conclude that x is 3 or greater. This means that x=3 and y=1, since we can now rule out the solution x=1 and y=3. SUFFICIENT

The answer is D

(B) If 121 (aka 11^2) is not a divisor of P, we know that y must be less than 2. In other words y must equal 1, in which case our only possible solution is x=3 and y=1. SUFFICIENT
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by Abdulla » Sun Nov 15, 2009 8:21 pm
Brent Hanneson wrote:
heshamelaziry wrote:Positive integer P has exactly 2 positive prime factors, 5 and 11. If P has a total of 8 positive factors, including 1 and P, what is the value of P?
A) 125 is a factor of P.
B) 121 is not a factor of P.
Great question.
First, there's a little rule you should know:
If the prime factorization of a number, n, is such that n = p^a * q^b * r^c etc where p, q, r (etc) are prime numbers, then the number of positive divisors (factors) that n has is (a+1)(b+1)(c+1) etc
(e.g., 600=2^3 * 3^1 * 5*2, so 600 has 24 positive divisors since(3+1)(1+1)(2+1)=24)

We are told that P has two unique prime factors (5 and 11), which means that we can write P as follows: P = 5^x * 11^y, where x and y are both positive integers.
Our goal here is to find the value of P. In other words, we need to find the values of x and y.

Since we are told that P has exactly 8 positive divisors, we know (from the above rule) that (x+1)(y+1)=8
Since x and y must be positive integers, there are only two possible sets of values that satisfy the equation (x+1)(y+1)=8. The two possible solutions are i) x=1 and y=3, and ii) x=3 and y=1.

(A) if 125 (aka 5^3) is a factor of P, then we can conclude that x is 3 or greater. This means that x=3 and y=1, since we can now rule out the solution x=1 and y=3. SUFFICIENT

The answer is D

(B) If 121 (aka 11^2) is not a divisor of P, we know that y must be less than 2. In other words y must equal 1, in which case our only possible solution is x=3 and y=1. SUFFICIENT
Brent you're brilliant ...
Abdulla
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by heshamelaziry » Sun Nov 15, 2009 10:04 pm
I do not understand the above solution. Seems too complicated one for number properties. Why should the factor be 5^x and 11^y ?
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by Abdulla » Sun Nov 15, 2009 10:31 pm
heshamelaziry wrote:I do not understand the above solution. Seems too complicated one for number properties. Why should the factor be 5^x and 11^y ?
Hesham,

There is a general rule to find the number of all factors of a certain number you MUST memorize .

For example,
what is the number of all factors of 2500?
when you factor it, you will get 5,5,5,5,2,2
which is 5^4 * 2^2
now simply add one to each exponent and then multiply them to get the answer.
(4+1)(2+1) = 5*3 = 15 ...

Just memorize the rule.
Abdulla
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by heshamelaziry » Sun Nov 15, 2009 10:50 pm
what is 15 ? I don't think my score will depend on this question. i don't need 720 instead of 700, so I don't need to know everything. i don't need to exhaust my time and nerves for this mf question. VOID
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by Brent@GMATPrepNow » Mon Nov 16, 2009 8:16 am
heshamelaziry wrote:what is 15 ? I don't think my score will depend on this question. i don't need 720 instead of 700, so I don't need to know everything. i don't need to exhaust my time and nerves for this mf question. VOID
Perhaps you're right, heshamelaziry; this is an advanced question.
Having said that, the main principle involved is not beyond the 650 level. You should be aware of the following rule:

If the prime factorization of a number, n, is such that n = p^a * q^b * r^c etc where p, q, r (etc) are prime numbers, then the number of positive divisors (factors) that n has is (a+1)(b+1)(c+1) etc
Example: 600=2^3 * 3^1 * 5^2, so 600 has 24 positive divisors since(3+1)(1+1)(2+1)=24
Brent Hanneson - Creator of GMATPrepNow.com
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by antondesh » Mon Nov 16, 2009 11:58 am
Hi Brent,

I'm a little unclear on your explanation. I know you say that the only two possible solutions for (x+1)(y+1)=8 are 1 and 3 but what about 0 and 7? You say that "x and y have to be positive integers", is that necessarily the case?

If I tried to work this problem through without knowing your method, this would be my thought process:

The problem tells us that the integer P has 8 positive factors INLCUDING 1 and P. So it really has 6 prime factors, which could either be a 5 or 11.

So factors of P = 1, P, 5, 11, [5 or 11], [5 or 11], [5 or 11], [5 or 11]

Statement A:
125 is a factor of P. This tells me that we know we have at least three 5's. So according to statement A, factors of P = 1, P, 5, 11, 5, 5, [5 or 11], [5 or 11]. I still don't know whether the last two factors are 5's, 11's or a combination of both. INSUFFICIENT.

Statement B:
121 is not a factor of P. This tells me that we only have one 11, the rest must be 5's. SUFFICENT.


Can you please explain where I'm making an error? Thanks.
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by okigbo » Tue Nov 17, 2009 10:09 am
Y cannot be 0 in (x+1)(y+1)=8. That would mean that 11 is not a prime factor in p which goes against the question stem. Correct me if I am wrong but x and y have to be at least 1.
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by antondesh » Tue Nov 17, 2009 11:18 am
I think I get it now - I was reading the problem wrong.

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

What it's basically saying is that the factors of P are 5, 11, 1 and P and there are also 4 other non-prime factors. So that nifty formula gives us the number of all positive factors of a number, not just the prime ones. Does that number include factors such as 1 and the number itself?

For example, let's take 500.

Prime factorization = 2^2 * 5^3

So this means 500 has a total of (2+1)*(3+1)= 12 factors? Does this number include 500 and 1 too?
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by anu8 » Tue Nov 17, 2009 12:13 pm
Hi can anyone please help me understand these number property DS problems?

this is my weakest area in math...


1) If x and y are positive, is x^3 > y?
(1) sq. rt (x) > y
(2) x > y

2)What is the value of the integer k?
(1) k + 3 > 0
(2) k^4 0
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