say n is a number expressed as a^x*b^y
where a and b are its prime factors raised to certain powers.
number of factors of n is given by (x+1)*(y+1)
(2^y)*(3^3)
=> number of factors: (y+1)*(3+1)=4*(y+1)
2^51 has 51+1=52 factors.
given that 4*(y+1)=52
=>y+1=13
=>y=12
hence B
factors
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
-
scoobydooby
- Legendary Member
- Posts: 1035
- Joined: Wed Aug 27, 2008 10:56 pm
- Thanked: 104 times
- Followed by:1 members
-
bluementor
- Master | Next Rank: 500 Posts
- Posts: 418
- Joined: Wed Jun 11, 2008 5:29 am
- Thanked: 65 times
You can find the numer of different factors for a number using the following procedure:
1. prime factorize the number
2. add 1 to the powers of the prime numbers.
3. multiple the added powers to get the number of different powers.
For example: find the number of different factors for 144.
1. prime factorize: 144=9x16 = (3^2)*(2^4)
2. add 1 to the powers of the prime numbers: (2+1) and (4+1)
3. multiple the added powers: (2+1)*(4+1) = 3x5 = 15 different factors for 144.
Using the same principle:
The number of different factors for 2^51 = 52.
The number of different factors for (2^y)*(3^3) = (y+1)(3+1)
and since (y+1)(3+1) = 52, then:
4(y+1) = 52
y+1 = 13
y = 12.
Choose B.
-BM-
1. prime factorize the number
2. add 1 to the powers of the prime numbers.
3. multiple the added powers to get the number of different powers.
For example: find the number of different factors for 144.
1. prime factorize: 144=9x16 = (3^2)*(2^4)
2. add 1 to the powers of the prime numbers: (2+1) and (4+1)
3. multiple the added powers: (2+1)*(4+1) = 3x5 = 15 different factors for 144.
Using the same principle:
The number of different factors for 2^51 = 52.
The number of different factors for (2^y)*(3^3) = (y+1)(3+1)
and since (y+1)(3+1) = 52, then:
4(y+1) = 52
y+1 = 13
y = 12.
Choose B.
-BM-
