babyZ wrote:While waiting in line at the supermarket, one shopper recognizes the lady in front of him as one of his neighbors. Being interested in finding a baby-sitter for his own child, he asks her about the ages of her three daughters. The woman replies to him that the product of their ages is 72 and the sum of their ages is the same as her house number. With her house number in mind, he replies, "That's not enough information. Give me another clue." The woman replies, "My oldest daughter has taken a baby-sitting class." With this additional clue, he is able to compute the ages of the three girls. What are their ages and how did he figure them out?

You know that their ages, when multiplied, give you 72. So first list all the factorizations that result in 72:

. . . . .1, 1, and 72

. . . . .1, 2, and 36

. . . . .1, 3, and 24

. . . . .1, 4, and 18

. . . . .1, 6, and 12

. . . . .1, 8, and 9

. . . . .2, 2, and 18

. . . . .2, 3, and 12

...and so forth.

Then note that it was not sufficient to know the neighbor's house number, so there must be factorizations which have the same sum. So add this to your listing:

. . . . .1, 1, and 72: 1 + 1 + 72 = 74

. . . . .1, 2, and 36: 1 + 2 + 36 = 39

. . . . .1, 3, and 24: 1 + 3 + 24 = 28

. . . . .1, 4, and 18: 1 + 4 + 18 = 23

. . . . .1, 6, and 12: 1 + 6 + 12 = 19

. . . . .1, 8, and 9: 1 + 8 + 9 = 18

. . . . .2, 2, and 18: 2 + 2 + 18 = 22

. . . . .2, 3, and 12: 2 + 3 + 12 = 17

...and so forth.

Once you've reduced your list to only those products whose factors sum to the same values (you may have more than one such set), use the other clue: the fact that there is an "oldest". In particular, this means that the three girls are not triplets and, if there are twins, they are the two younger children.

Have fun!