View Full Version : Math Problem

Jim Brier
09-08-2002, 05:50 PM
A census taker asks a woman "How many children do you have?". She says "Three". The census taker asks "What are their ages?". She responds, "The product of their ages is 36 and the sum of their ages is the house number next door." The census taker walks next door, reads the house number, appears puzzled, and returns to the woman, asking: "Is there something you forgot to tell me?" The woman responds: "Yes, the oldest child is in the park."

How old is each child?

09-08-2002, 06:34 PM
Great question, Jim.

If the product is 36, the possibilities are:
36, 1, 1 (sum of 38)
18, 2, 1 (sum of 21)
12, 3, 1 (sum of 16)
9, 2, 2 (sum of 13)
6, 6, 1 (sum of 13)
6, 3, 2 (sum of 11)
4, 3, 3 (sum of 10)

The census taken wouldn't have had to go back for more information unless the house next door was #13. Since the woman said the oldest "child" instead of the oldest "children" the ages must be 9, 2, and 2 instead of 6, 6, and 1.

Jim Brier
09-08-2002, 10:18 PM
The perfect answer. One minor quibble. There are 8 possibilities, not 7. Ages of 9, 4, and 1 are also factors of 36 with a sum of 14. Of course, this number would not have confused the census taker either.

09-09-2002, 10:27 AM
One minor quibble. There are 8 possibilities, not 7.

Sorry. I'll try harder next time. /forums/images/icons/wink.gif

Ed Miller
09-10-2002, 05:16 AM
The children are 9, 2, and 2. You are looking for a number X such that a*b*c = 36 and a+b+c = X AND d*d*e = 36 and d+d+e = X where d > e (i.e. there is no oldest child). If you let a=9, b=2, c=2, d=6, e=1, these are satisfied for X=13.

Jim Brier
09-10-2002, 12:28 PM
How is it determined that two of three ages must be the same at the start of the problem? Using your methodology should it not be d*e*f = 36 and d + e + f = X? How can you immediately eliminate one of the variables? In fact, at this point in the formulation of the problem, how is it known that there are not other sets of three factors whose product is 36 that add up to X?

John Feeney
09-10-2002, 02:48 PM
Don't try harder, Uston; for you tried exactly hard enough. Embrace that.

09-10-2002, 07:08 PM
I'll make a deal with you. I won't try harder if you'll write another book. Your first one is, in ghetto parlance, the shiznit. /forums/images/icons/smile.gif

Ed Miller
09-10-2002, 09:19 PM
I eliminate one of the variables because the "oldest child" thing implies that one solution has an oldest child while another does not (hence, the "oldest" and "middle" children must be the same age). I can't immediately eliminate the variable.. I only do that at the end when we get the "oldest child" information.

As for why I know that there are not other sets of three factors whose product is 36 and that add up to X, if there were, then the problem would not have a unique answer. I am assuming, because you ask the question, that there is a unique answer.

John Feeney
09-11-2002, 02:15 AM
Thanks for the props, man. If I get back into poker I may think about another book. But it's a big world out there. I didn't even get to Rangoon this summer. /forums/images/icons/wink.gif

bad beetz
09-11-2002, 07:31 PM
I agree, you book is the shiznitobam schnip snap sack.

John Feeney
09-11-2002, 09:09 PM