If a and b are positive integers, and

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

Can someone explain this question in detail?

Please re-read the 2-page discussion there is and ask more specific question. Thank you.

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

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.

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

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