Exponents-

(2^3)(3^4)(5^7) = b*(a^3)

b = (2^3)(3^4)(5^7)/a^3

Value of A = 1, 2, 3, 5, 25(5x5), 6(2x3), 10(2x5), 50(2x5x5), 15(3x5), 45(3x5x5), 30(2x3x5), 150(2x3x5x5)

Correspondingly the value of "b" will change.

So, 12

Answer [spoiler]{E}[/spoiler]

theCodeToGMAT wrote:(2^3)(3^4)(5^7) = b*(a^3)

b = (2^3)(3^4)(5^7)/a^3

Value of A = 1, 2, 3, 5, 25(5x5), 6(2x3), 10(2x5), 50(2x5x5), 15(3x5), 45(3x5x5), 30(2x3x5), 150(2x3x5x5)

Correspondingly the value of "b" will change.

So, 12

Answer [spoiler]{E}[/spoiler]

Rahul,
Can you post any other method ? (Since we may miss any one of the possibility during exam time)
or Instead writing down all possible options is there a way to find 12 possibilities from the numbers 2,3,5.

Thanks in advance

Regards,
Uva.

Hi All,
If the above question has been restated with a as Prime number then can we use the below method ?
(2^3)(3^4)(5^7) = b*(a^3)

In LHS you can see there are (3+1)*(4+1)*(7+1) = 4*5*8 factors

In RHS you can see (3+1)*b =4*b factors

So b can take 40 factors

so b can take 40 different possible values

Please help me whether I am wrong or right.

Thanks in advance.

Regards,
Uva.

~~mistakenly posted~~

theCodeToGMAT wrote:

Uva@90 wrote:Hi All,
If the above question has been restated with a as Prime number then can we use the below method ?
(2^3)(3^4)(5^7) = b*(a^3)

In LHS you can see there are (3+1)*(4+1)*(7+1) = 4*5*8 factors

In RHS you can see (3+1)*b =4*b factors

So b can take 40 factors

so b can take 40 different possible values

Please help me whether I am wrong or right.

Thanks in advance.

Regards,
Uva.

See UVa, you cannot assume "a" & "b" as Prime Number..

Rahul
I am asking , if in question if they had mentioned Explicitly as Prime number then is my approach correct ?

[email protected] wrote:If a and b are positive integers, and (2^3)(3^4)(5^7) = (a^3)*b, how many different possible values of b are there?
(A) 2
(B) 3
(C) 4
(D) 6
(E) 12

Ans E

(2^3)(3^4)(5^7) = (a^3)*b

Here, since all the factors have powers greater than or equal to 3, 'a' can be any one number or a combination of the 3.

Total possibilities:
None of the given factors (1^3): 1
One at a time: 3
Two at a time: 3C2 + 1 = 3 + 1 = 4
Three at a time: 3C3 + 2C1 = 1 + 3 = 3
Four at a time: 1

Total = 1 + 3 + 4 + 3 + 1 = 12

Choose E

Cheers

Another way of doing this is just grouping the terms into threes.

(2 * 2 * 2) (3 * 3 * 3) (5 * 5 * 5) (5 * 5 * 5) * 3 * 5

We have four different "cubes" here, so we could choose 0 to 4 of them - but be careful, as two of them are identical! (If they weren't, the problem would have a nice shortcut.)

Now we just need to decide how many of these "cubes" we want in our a. Each of these are small enough that you're probably best just counting out each case. (For "2 of them", for instance, I would chose 2+3, 2+5, 3+5, or 5+5.)

0 of them: 1 way
1 of them: 3 ways
2 of them: 4 ways
3 of them: 3 ways
4 of them: 1 way

And we're done!