Question 275191
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?
I understand that her brothers are twins, so two of the three numbers I'm looking for are the same, but beyond that, I dont really know what to do. 

Was there no other information about Kiana's age? 

If not,let x be the age of each twin. Then we need to look at the values of x^2 less than 128. So we have:

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81
10^2 = 100
11^2 = 121
12^2 = 144

The ages of the two boys must be less 11 or less. We know when we multiply one of these square values by Kiana's age we get 128 and Kiana's age must be less than the age of the twins.  The squares above that divide evenly into 128 are 1, 4, 16 and 64. In the first three cases Kiana's age would need to be greater that the twins. For the case where the twins are both 8 and the product of their ages is 64 Kiana's age would need to be 128/64 = 2. So we have the twins each age 8 and Kiana age 2. The sum then is 8 + 8 + 2 = 18.