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If a and b are positive integers, and

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mikemcgarry

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10 Feb 2015, 10:36

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mike34170

Don't know whether i've been lucky with that but i handled it this way :

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

Number of possible different factors for (a^3)*b =
(2^3) = 3 (i.e = 2; 4; 8)
(3^4) = 4 (i.e = 3; 9; 27; 81)
(5^7) = 7 (i.e = 5; 25; 125; 725; Etc.)
And don't forget 1 that is a common factor to each of the expressions.

So we have 3 + 4 + 7 + 1 = 15 different factors for (a^3)*b.
Then (a^3) could embrace 3 of these factors
So b could be 15 - 3 = 12 different values

Magoosh what do you think about it ?

Dear mike34170,
First of all, since we have the same name, you might as well call me by name. :-)

I'm happy to respond. :-)

Here's the deal. This problem, that I created on a whim, is ultimate a faulty question, because there are a number of invalid ways to get to the correct numerical answer. The answer is 12, but the way you have approached it is incorrect. Among other things, when task #1 can be done in N1 ways, and independently, task #2 can be done in N2 ways, we don't add---we multiply! That's the Fundamental Counting Principle. See:
https://magoosh.com/gmat/2012/gmat-quant-how-to-count/
Also, seem to be counting what would be possible for b, without taking into account the special constraint that a must be cubed.
Also, 5^4 = 625, not 725.

See summer101's post above, which contains a significantly better question. If you understand that and its solution, you really understand this material.

Mike :-)

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08 Mar 2016, 06:26

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mikemcgarry

I was inspired by another, easier question to create this question, just for fun.

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

You may find this blog on prime factors helpful in thinking about this problem:
https://magoosh.com/gmat/2012/gmat-math-factors/
You may also find this blog on counting helpful.
https://magoosh.com/gmat/2012/gmat-quant-how-to-count/

I will post a full solution if there's interest.

Excellent question mike
I wish i could give you more than one kudos on this one..

i got to 11 cases hence i chose E
but the big takeaway from this was a being one and the complete product being b

Regards

vitaliyGMAT

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14 Nov 2016, 11:19

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mikemcgarry

Nice question and great job! It really encompass different fields of math.

I’m too late but here is my view for the sake of diversity.

I prefer to solve it for \(a^3\) rather than \(b\) as \(a^3\) is the most restrictive part.

Let’s create different cases for different perfect cubes.

1. \(a^3\) consists of a single prime – \(p^c\), where \(c\)- multiple of 3.
We have \(2^3\), \(3^3\), \(5^3\) and \(5^6\)

\(3C1 + 1 = 4\)

2. \(a^3\) is composed from two distinct primes – \(p^cq^d\), where \(c,d\) – multiples of 3.

\(2^33^3, 2^35^3, 3^35^3\) and \(2^35^6, 3^35^6\)

\(3C2 + 1*2C1 = 5\)

3. \(a^3\) – 3 distinct primes – \(p^cq^dr^e\), where \(c,d,e\) – multiples of 3.

\(2^33^35^3\) and \(2^33^35^6\) together \(= 2\)

4. and final case when all primes will be absorbed by \(b\) – \(1\) (also a perfect cube) \(p^0q^0r^0\) = 1

Total # = \(4 + 5 + 2 + 1 = 12\)

Fdambro294

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16 Jan 2021, 01:39

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Since the highest count is 12, we can "Brute Force" the counting and take it on a case-by-case method:

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

whichever Prime Bases we choose to fill the (a)^3 = a * a * a spot

then the Value of (b) will automatically be determined.

so we can count the different values of (b) by determining in how many ways we can fill the (a)^3 Factor

Limiting Constraint: if we place at least 1 Prime Base into (a) ------> we must place a total of 3 of that Prime Base into (a)^3

Case 1: Only 1 Prime Base is Used for (a)

(a)^3 = a * a * a

2 * 2 * 2

3 * 3 * 3

5 * 5 * 5

(5 * 5 * 5) (5 * 5 * 5)

4 Values that can fill (a)^3 -----> and (b)

Case 2: 2 Prime Bases are used for (a)

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

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

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

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

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

5 Different Values that can fill (a)^3 ----- which means 5 different values that can fill (b)

Case 3: All 3 Prime Bases are used

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

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

Last Case is EVERY Prime Base in (b) ------ and (a) = 1

3 Values

4 + 5 + 3 = 12

(E)

luckily, we could have stopped once we crossed (D)'s threshold

fantastic problem

Karnk

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11 Jun 2022, 21:05

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here is an easy way to solve it.
we can first find out how many cubes can be formed. lets c...2^3, 3^3,5^3,10^3,15^3,6^3,30^3 alrdy 7..so any answer more thn 7 in options is 12.

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13 Oct 2022, 05:27

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mikemcgarry

1) Number of different possible values of b is equal to number of different possible values of a.
2) Number of different possible values of a
= [no. of possible combinations of (2^3)(3^4)(5^3)] + [no. of possible combinations of (2^3)(3^4)(5^6)]
= 6 + 6
=12

E is Correct

luisdicampo

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05 Dec 2025, 08:12

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mikemcgarry

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

Show Spoiler

I was inspired by another, easier question to create this question, just for fun.

You may find this blog on prime factors helpful in thinking about this problem:
https://magoosh.com/gmat/2012/gmat-math-factors/
You may also find this blog on counting helpful.
https://magoosh.com/gmat/2012/gmat-quant-how-to-count/

I will post a full solution if there's interest.

Deconstructing the Question
We are given the equation:
\((2^3)(3^4)(5^7) = a^3 \cdot b\)
Where \(a\) and \(b\) are positive integers.
We need to find the number of possible values for \(b\).

Rearranging the equation:
\(b = \frac{2^3 \cdot 3^4 \cdot 5^7}{a^3}\)

Since \(b\) must be an integer, \(a^3\) must be a factor of \(2^3 \cdot 3^4 \cdot 5^7\).
This means \(a\) must be formed by prime factors \(2, 3, 5\) such that its cube fits within the given exponents.

Let \(a = 2^x \cdot 3^y \cdot 5^z\).
Then \(a^3 = 2^{3x} \cdot 3^{3y} \cdot 5^{3z}\).

Step-by-Step Analysis of Exponents

1. Prime Factor 2: Available power is \(2^3\).
We need \(3x \le 3 \implies x \le 1\).
Possible values for \(x\): 0, 1.
(2 options)

2. Prime Factor 3: Available power is \(3^4\).
We need \(3y \le 4 \implies y \le 1.33\).
Possible values for \(y\): 0, 1.
(2 options)

3. Prime Factor 5: Available power is \(5^7\).
We need \(3z \le 7 \implies z \le 2.33\).
Possible values for \(z\): 0, 1, 2.
(3 options)

Calculation
Total possible combinations for \(a\) (and therefore for \(b\)) is the product of the options:
\(Total = 2 \times 2 \times 3 = 12\).

Answer: E

ManishaPrepMinds

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26 Dec 2025, 04:42

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Given:
2^3 · 3^4 · 5^7 = a^3 · b
with a and b positive integers.
Only exponents that are multiples of 3 can go into a^3, the remaining goes into b.
For each prime:
2^3 → multiples of 3 possible: 0, 3 → 2 choices
3^4 → multiples of 3 possible: 0, 3 → 2 choices
5^7 → multiples of 3 possible: 0, 3, 6 → 3 choices
Each choice of exponents used in a^3 leaves a unique remaining factor, so each choice gives a different value of b.
Total possible values of b:
2 × 2 × 3 = 12

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