Math
posted by Sarah .
The census taker says, "I need to know the ages of your children."
The mother replies, "I have no oneyearolds. The product of my
children's ages is 90, and the sum of their ages is the same as my
house number."
The census taker replies. "I can see the house number but I still need
more information."
The mother says, "You're right. You also need to know that the boy
across the street is older than my oldest child."
The census taker says, "Thank you, I now know the ages of your
children."
What are the ages of the children?
What is the house number?
What is the age of the boy across the street?
I made a list of factors, but don't know where to go from there. Thank you for your help!

Can you post the list of factors that you have made?

I feel like I'm missing information, but here's the best I can do.
Product of ages = 90
Prime factors of 90: 2,3,3,5
So the possible number and age of children are:
2,3,3(twins),5
3,5,6
2,5,9
3,3(twins),10
2,3,15
6,15
5,18
Since the mother said "...older than my oldest child.", there must be three or more children, or even three different ages. That leaves us with:
2,3,3(twins),5
3,5,6
2,5,9
2,3,15
Some cities do not put 13 as a house number, so we eliminate the first case, to leave us with:
3,5,6 (sum=14)
2,5,9 (sum=16)
2,3,15 (sum=20)
If we eliminate ages that are too close, one year apart (but logically & physically feasible), that leaves us only with 2,5,9 (sum=16).
Also, the boy across the street can be one or more years older than the oldest child.
If you have other hints/conclusions, please post.