factors

[figs]

If the number of different positive factors of (2^y)*(3^3) is the same as the number of different factors of (2^51) what is the value of y?

a 11
b 12
c 13
d 48
e 51

can anybody help?
OA after explanation

[scoobydooby]

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

[bluementor]

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-

[figs]

very clear
OA B