I need help with this math problem: How many positive vaules for A+B+C, if (A)(B)(C)=72?

I'm assuming A, B, and C must be positive integers.





72


= 2 * 36


= 2 * 2 * 18


= 2 * 2 * 2 * 9


= 2 * 2 * 2 * 3 * 3





So you can have


2 * 2 * (2 * 3 * 3) = 2 * 2 * 18


2 * (2 * 2) * (3 * 3) = 2 * 4 * 9


(2 * 2) * (2 * 3) * 3 = 4 * 6 * 3


(2 * 2 * 2) * 3 * 3 = 8 * 3 * 3


2 * (2 * 2 * 3) * 3 = 2 * 12 * 3


(2 * 3) * (2 * 3) * 2 = 6 * 6 * 2





That's 6 possibilities for products.





Now just make sure the sums are different:


2 + 2 + 18 = 22


2 + 4 + 9 = 15


4 + 6 + 3 = 13


8 + 3 + 3 = 14


2 + 12 + 3 = 17


6 + 6 + 2 = 14 (duplicate)





So there are 5 possibilities for sums.

I need help with this math problem: How many positive vaules for A+B+C, if (A)(B)(C)=72?
That first answer's wrong as A, B and C are different,


2+2+18 won't do neither will 8+3+3.


So there are only 3 possibilities. Report It

Reply:Also, the answer neglects situations where two of the numbers are negative, but the sum is still positive. For instance, -2 * -4 * 9 = 72 and 9-2-4=3 Report It

Reply:There is an infinite number of ways to multiply 3 positive numbers and get 72. Thus there is an infinite number of values for A+B+C