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

Sir, how did you do that?

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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.

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

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.

wow

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

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

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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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) = [(2^r)(3^s)(5^t)]^3 x b

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

b = [2^(3 - 3r)] x [3^(4 - 3s)] x [5^(7 - 3t)]

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.

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

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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