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If a and b are positive integers, and (2^3)(3^4)(5^7)

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by GMATGuruNY » Sat Aug 03, 2013 1:21 pm
guerrero 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

OA E
(2³)(3³)(5³)(5³)(3)(5) = (a³)b

a³ must be the cube of an integer
Once a value has been chosen for a³ , the value of b will be equal to whichever factors on the lefthand side are not included in the value of a³.

The answer choices represent the number of options for b.
Since the greatest answer choice is 12, there are at most 12 options for b, implying that there are at most 12 options for a³.
Thus, we can quickly count the number of options for a³:
1³
2³
3³
5³
(2³)(3³)
(2³)(5³)
(3³)(5³)

We can stop here.
Since there are at least 7 options for a³, there are at least 7 options for b.
Thus, the only viable answer choice is E.

Here are all of the options for a³ :
1³
2³
3³
5³
(2³)(3³)
(2³)(5³)
(3³)(5³)
(5³)(5³)
(2³)(3³)(5³)
(2³)(5³)(5³)
(3³)(5³)(5³)
(2³)(3³)(5³)(5³)
Total options = 12.
Last edited by GMATGuruNY on Mon Aug 05, 2013 10:59 am, edited 1 time in total.
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by Matt@VeritasPrep » Sat Aug 03, 2013 1:47 pm
Another approach:

Once we have the equation

(2³)(3³)(25³)(3)(7) = (a³)b

We can say that whatever a is, its prime factors must be some combination of (2 * 3 * 25), as it can only take factors that are "cube-able", to coin a phrase. (For example, a can't contain a 7, as we don't have three 7's to put in a * a * a.)

So the real question is "how many factors does 2*3*25" have, as each of the factors of this number are possible values of a. 2 * 3 * 5² has (1+1)*(1+1)*(2+1) factors, using the formula for the # of factors, so there are 12 unique factors, each of which can be the value of a.
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by ela07mjt » Sat Aug 03, 2013 2:47 pm
"Once a value has been chosen for a³ , the value of b will be equal to whichever factors on the lefthand side are not included in the value of a³".

Then how there are 12 different possible values of b. How does calculating factors for a tells you possible values of b?
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by faraz_jeddah » Sun Aug 04, 2013 5:47 am
GMATGuruNY wrote:
(2³)(3³)(5³)(5³)(3)(7) = (a³)b
Ignoring the typo in the solution you posted Mitch can you please tell me all the possible values for b as you have listed the possible values of a.

My approach-

The equation can be written as

(2³)(3³)(5³). 3 . 5^4 = (a³)b

Since a has to be a perfect cube, I can list out the possible values of b

3
3 .(5)
3 .(5^2)
3 .(5^3)
3 .(5^4)
5
5^2
5^3
5^4

Thats 9 Values.


Again the equation can be rewritten as

(2³)(3³)(5^6). 3 . 5 = (a³).b

again I can list out possible values of b as

3
3.5
5

Thats 3 values

Total values 9 + 3 = 12

Is this approach correct?
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by GMATGuruNY » Mon Aug 05, 2013 11:29 am
faraz_jeddah wrote:
My approach-

The equation can be written as

(2³)(3³)(5³). 3 . 5^4 = (a³)b

Since a has to be a perfect cube, I can list out the possible values of b

3
3 .(5)
3 .(5^2)
3 .(5^3)
3 .(5^4)
5
5^2
5^3
5^4

Thats 9 Values.


Again the equation can be rewritten as

(2³)(3³)(5^6). 3 . 5 = (a³).b

again I can list out possible values of b as

3
3.5
5

Thats 3 values

Total values 9 + 3 = 12

Is this approach correct?
Perfect reasoning. Nicely done!
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