Can you guess their ages?

The first answer given by Mr. Alfred was that the product of his sons' ages is 36. Hearing this, genius Dr, Bertrand factorized 36 into triplets in all possible ways in his brain. The following were his results: $1\times 1\times 36=36\qquad 1\times 2\times 18=36\qquad 1\times 3\times 12=36\\ 1\times 4\times 9=36\qquad \quad 1\times 6\times 6=36\qquad \quad 2\times 2\times 9=36\\ 2\times 3\times 6=36\qquad \quad 3\times 3\times 4=36$ At this stage, Dr. Bertrand was not able to decide anything when he got his next clue : their ages add up to his house number. Dr. Bertrand's abilities enabled him to add all the triplets and mentally list down the sums. Now think it in this way : had Dr. Bertrand found a unique sum matching with his house number, he would have guessed the ages correctly at this stage. But he could not do so. This means at least two of the above triplets add up to the same number, We perform the following computation: $1+1+36=38\qquad 1+2+18=21\qquad 1+3+12=16\\ 1+4+9=14\qquad \quad 1+6+6=13\qquad \quad 2+2+9=13\\ 2+3+6=11\qquad \quad 3+3+4=10$ and see that there are exactly two triplets that add up to 13. So, the house number must be 13 and the ages must be one of these two triplets. The third answer says something about the eldest son. Since the ages are integers, (1,6,6) gan't be a possibility as then there will be two eldest sons. So the ages are 2,2,9. And $\\ { 2 }^{ 2 }+{ 2 }^{ 2 }+{ 9 }^{ 2 }=89$ .

We know that 36 = 3x3x2x2.

Since he has one eldest son, and Bertrand received useful information upon this revelation, one of the possibilities in his head must have been one in which there are two eldest sons.

We see that in order to have two sons of identical age, we must maintain two of the identical factors and multiply the other two, or multiply both factors and set the third age as 1.

Maintaining two factors and multiplying the others yields either 4,3,3 or 9,2,2. But neither of these possibilities present two oldest sens. Ergo the possibility bothering Bertrand was 6,6,1.

Having established this as something Bertrand was thinking about, we can conclude that his house number is 6+6+1=13.

Since his number is 13, and the guy has 1 eldest son, 4,3,3 is ruled out (sum is 11), 6,6,1 is ruled out (two eldest), and 9,2,2 remains. Sum of squares is 89.