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

Question

If a and b are positive integers, and  (2^3)(3^4)(5^7) = a^3b , how many different possible values of b are there?

A. 2
B. 4
C. 6
D. 9
E. 12

GMATinsight · Accepted Answer

ShankSouljaBoi

basically we need to understand that  (2^3)(3^4)(5^7) = a^3b  is expecting from us to separate a cube from the rest of the number

so we should be able to write the left part of the equation in the form that is represented on the right side of the equation

Let,  2^3 = p 
and  3^3 = q 
and  5^3 = r

So we have  p*q*r^2  to represent  a^3

Total factors of  p*q*r^2  = (1+1)*(1+1)*(2+1) = 12

i.e. there are 12 ways to represent  a^3  and the remaining part of the expression on left represents  b

hence, 12 cases

Answer: Option E

eswarchethu135 · Answer

Total no. of possible values is 12.

1)  2^3 (3^4 * 5^7) 
2)  3^3 (2^3 * 3 * 5^7) 
3)  5^3 (2^3 * 3^4 * 5^4) 
4)  (5^2)^3 (2^3 * 3^4 * 5) 
5)  (2*3)^3 (3 * 5^7) 
6)  (2*5)^3 (3^4 * 5^4) 
7)  (3*5)^3 (2^3 * 3 * 5^4) 
8)  (2*{5^2})^3 (3^4 * 5) 
9)  (3*{5^2})^3 (2^3 * 3 * 5) 
10)  (2*3*5)^3 (3 * 5^4) 
11)  1^3 (2^3 * 3^4 * 5^7) 
12)  (2*3*{5^2})^3 (3 * 5)

OPTION :  E

Shrey9 · Answer

Sir, how did you do that?

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eswarchethu135 · Answer

For me it's intutive. I have past experiences on how to respond for these kind of problems. I don't have any strategy to tell. But only practise can get you arrive solutions for these questions.

Posted from my mobile device

ShankSouljaBoi · Answer

Any other way than actually calculating by brute force ?
 VeritasKarishma   GMATinsight

VeritasPrepBrian · Answer

It's slightly brute-force but I'd look at it this way:

-Of the exponents given, you can break out four sets of "prime base to the third" -- 2^3, 3^3 (leaving 3^1 behind), 5^3, and 5^3 (since you have 5^7, you have two sets of 5^3 you can use)

-Then there are four types of combinations you can use as your a^3: one of the four (e.g. 2^2), two of the four (e.g. 2^2 * 5^3), three of the four (e.g. 2^2 * 3^3 * 5^3), or all four.

Then you just need to remember that the 5^3s repeat, so you have the options of:

All four --> one way to do it

Three of the four --> 2, 3, and 5; 2, 5, and 5; and 3, 5, and 5 --> three ways to do it

Two of the four --> 2 and 3; 2 and 5; 3 and 5; 5 and 5 --> four ways to do it

One of the four --> the 5s repeat, so you could use 2, 3, and 5 --> three ways to do it

That gets you to 11 and you know they're all valid, so you can use the answer choices to say that you have to be missing one somewhere and pick 12. *Or* you can have the presence of mind to realize that 1^3 works as a^3 and the rest could all be part of b. Honestly...I don't think I'd see that up front but the answer choices here would definitely guide me to that, or if 11 and 12 were each options hopefully I'd do that "hey am I missing one?" double-check.

This is kind of brute-force, but I think at least organized enough that it's replicable.

ShankSouljaBoi · Answer

wow

ScottTargetTestPrep · Answer

To determine possible values of b, we have to explore all the possible values of a.

We see that a can be of the form 2^r x 3^s x 5^t, where r is 0 or 1, s is 0 or 1 and t is 0, 1, or 2, so that b is still an integer.

Since the values of r and s have 2 choices each and the value of t has 3 choices, there are a total of 2 x 2 x 3 = 12 different values for a, and, hence, there are also a total of 12 different values for b.

Answer: E

Manfo · Answer

I would suggest the below approach, not too much theory, not too much brute force..

Keep in mind that both  a  and  b  are positive and integers, therefore if we move  a  to the denominator it gives us:

\frac{(2^3)(3^4)(5^7)}{(a^3)}  =  b

so we need to find those values of  a  for which  b  remains an integer:

(1) It's easy to find that  a  can take values like 1, 2, 3, 5 and  5^2

(2)  a  can take also values that result from the multiplications of the previous values ( 2*3 ), ( 2*5 ), ( 3*5 ) and ( 2*3*5 ) since at the numerator we have a multiplication of all three elements to the power of, at least, 3

(3) continuing the multiplication approach,  a  can also take values ( 5^2  *3 ), ( 5^2  *2 ) and ( 5^2  *3*2 ), since at the numerator the number 5 has power 7

Total number of values that  a  can take is 12

E

subramanya1991 · Answer

Hello Scott. Can you kindly explain how you got 3 choices for 5. I understood the 2*2 part of it but got held by by the 3.

Thanks

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ScottTargetTestPrep · Answer

Response:

Let’s substitute a = 2^r x 3^s x 5^t in (2^3)(3^4)(5^7) = (a^3)b:

(2^3)(3^4)(5^7) =  ^3 x b

(2^3)(3^4)(5^7) = (2^3r)(3^3s)(5^3t)b

b =   x   x

If b is to be an integer, each of the exponents 3 - 3r, 4 - 3s and 7 - 3t must be positive integers. For r and s, the only values which make the corresponding exponents positive integers are 0 and 1, but for t, the expression 7 - 3t will be a positive integer if t = 0, 1, or 2. That’s the reason there are three possible values for the exponent of 5.

Kinshook · Answer

Asked: If a and b are positive integers, and  (2^3)(3^4)(5^7) = a^3b , how many different possible values of b are there?

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

(a, b) =  { (2,3^4*5^7) ,  (3, 2^3*3*5^7) ,  (5, 2^3*3^4*4^4) ,  (5^2, 2^3*3^4*5) ,  (2*3, 3*5^7) ,  (3*5, 2^3*5^4) ,  (3*5^2, 2^3*5) ,  (2*5,3^4*5^4) ,  (2*5^2,3^4*5) ,  (2*3*5,3*5^4) ,  (2*3*5^2, 3*5) ,  (1,2^3*3^4*5^7) }; 12 values

IMO E

Moghrafu · Answer

2³3⁴ 5^7

When we choose how many different possible values of b, we will choose a -> 3 times from 2³ OR a-> 3 times from 3⁴ OR a-> 3 2times from 5^7

Hence,

³C³ + ⁴C³ + 7C6 = 1 + 4 + 7 = 12

The answer is: E

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siddhantvarma · Answer

There's a much simpler solution to this problem that doesn't involve any brute force approach. Using permutations, we can figure out the number of possible values b can take.

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

If you find the number of values a can take, the same number of values b can take. 
So we need to find the number of ways a can take unique values.

We know a is an integer, so when you pick numbers from the LHS you'll need to pick in cubes. Let's take 2^3. We can either take 2^0 (the case where we don't take 2 at all) or 2^3 so there are 2 ways to make a using 2^3.

Similarly, we have two ways to make a using 3^3. Either we take 3^3 or we don't take it at all.

For 5^7, we have three ways. One, we don't take 5 at all. Second, we take 5^3, and third, we take 5^6. Remember since we have 5^7 we can take 5^6 and still have a 5 left for b.

So total number of ways to make a: 2*2*3 = 12
These are the unique values a will take, and for each value of a that's formed, when you divide (2^3)(3^4)(5^7)/a you'll get a unique value of b. Hence b can be made in 12 ways.

Answer: E. 12

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

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

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