OT Puzzle: How old are your children?

primelord · 12-31-2003, 01:16 PM

Ok one more puzzle for today. I don't come across ones that I think will be challenging for the FOFC group very often so when I do I like to post them quickly.

Disclaimer: Again this puzzle was not created by me, but rather is just being reposted for your enjoyment.

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?

korme · 12-31-2003, 01:27 PM

2
3
6

primelord · 12-31-2003, 01:29 PM

Quote:

Originally posted by Shorty3281
2
3
6

Incorrect.

Huckleberry · 12-31-2003, 01:29 PM

2
2
9

There aren't another 3 ages that have the product of 36 and the sum of 11. In addition to 2+2+9=13 there is 1+6+6=13. So the next door house must have had the house number of 13 or else the census taker wouldn't have had to come back and ask for more information.

I think I worked this backwards but that should be right.

primelord · 12-31-2003, 01:34 PM

Ding! Ding!

Very good Huckleberry!

Draft Dodger · 12-31-2003, 01:41 PM

why doesn't Shorty's answer work?

primelord · 12-31-2003, 01:42 PM

Because 11 wasn't the number of the house next door.

Draft Dodger · 12-31-2003, 01:43 PM

primelord · 12-31-2003, 01:50 PM

Ok I guess I can give a more complete answer than that.

The key to the puzzle is the oldest child is in the park comment.

There are eight different ways to get a product of 36 from 3 numbers.

1. 36, 1, 1 (sum = 38)
2. 18, 2, 1 (sum = 21)
3. 12, 3, 1 (sum = 16)
4. 9, 4, 1 (sum = 14)
5. 9, 2, 2 (sum = 13)
6. 6, 6, 1 (sum = 13)
7. 6, 3, 2 (sum = 11)
8. 4, 3, 3 (sum = 10)

So if the house number had been anything but 13 then the census taker would have known each of their ages. However since the house number was 13 and there are two different ways to get to 13 the census taker had to go back for more information.

When the woman said the oldest "child" was in the park that clearly points to the 9, 2, 2 solution. If they were 6, 6, and 1 then she would have said oldest "children".

Draft Dodger · 12-31-2003, 01:57 PM

i get it now

Danny · 01-01-2004, 05:51 AM

Quote:

When the woman said the oldest "child" was in the park that clearly points to the 9, 2, 2 solution. If they were 6, 6, and 1 then she would have said oldest "children". [/b]

Not so, just because they are both 6, doesnt mean one is not older.

EagleFan · 01-01-2004, 11:32 AM

True, there is a loophole. It's possible, though not probable, to have 2 children that are 6 and not twins. Or the mother could be referring to the one that it 6, but minutes older, as the oldest.

Sorry, just had to play devil's advocate.

12-31-2003, 01:29 PM	#4
Huckleberry College Starter Join Date: Dec 2001	2 2 9 There aren't another 3 ages that have the product of 36 and the sum of 11. In addition to 2+2+9=13 there is 1+6+6=13. So the next door house must have had the house number of 13 or else the census taker wouldn't have had to come back and ask for more information. I think I worked this backwards but that should be right. __________________ The one thing all your failed relationships have in common is you. The Barking Carnival (Longhorn-centered sports blog) College Football Adjusted Stats and Ratings Last edited by Huckleberry : 12-31-2003 at 01:31 PM.

12-31-2003, 01:34 PM	#5
primelord Pro Rookie Join Date: Oct 2000	Ding! Ding! Very good Huckleberry! __________________ .

12-31-2003, 01:41 PM	#6
Draft Dodger Coordinator Join Date: Jan 2001 Location: Keene, NH	why doesn't Shorty's answer work? __________________ Mile High Hockey

12-31-2003, 01:42 PM	#7
primelord Pro Rookie Join Date: Oct 2000	Because 11 wasn't the number of the house next door. __________________ .

12-31-2003, 01:43 PM	#8
Draft Dodger Coordinator Join Date: Jan 2001 Location: Keene, NH	oh __________________ Mile High Hockey

12-31-2003, 01:27 PM	#2
korme Go Reds Join Date: May 2001 Location: Bloodbuzz Ohio	2 3 6

12-31-2003, 01:50 PM	#9
primelord Pro Rookie Join Date: Oct 2000	Ok I guess I can give a more complete answer than that. The key to the puzzle is the oldest child is in the park comment. There are eight different ways to get a product of 36 from 3 numbers. 1. 36, 1, 1 (sum = 38) 2. 18, 2, 1 (sum = 21) 3. 12, 3, 1 (sum = 16) 4. 9, 4, 1 (sum = 14) 5. 9, 2, 2 (sum = 13) 6. 6, 6, 1 (sum = 13) 7. 6, 3, 2 (sum = 11) 8. 4, 3, 3 (sum = 10) So if the house number had been anything but 13 then the census taker would have known each of their ages. However since the house number was 13 and there are two different ways to get to 13 the census taker had to go back for more information. When the woman said the oldest "child" was in the park that clearly points to the 9, 2, 2 solution. If they were 6, 6, and 1 then she would have said oldest "children". __________________ .

12-31-2003, 01:57 PM	#10
Draft Dodger Coordinator Join Date: Jan 2001 Location: Keene, NH	i get it now __________________ Mile High Hockey

01-01-2004, 11:32 AM	#12
EagleFan Hall Of Famer Join Date: Nov 2000 Location: Mays Landing, NJ USA	True, there is a loophole. It's possible, though not probable, to have 2 children that are 6 and not twins. Or the mother could be referring to the one that it 6, but minutes older, as the oldest. Sorry, just had to play devil's advocate. __________________ The EagleFan YouTube Channel - Reactions Soarin' & Scratchin' YouTube Channel - We react to music, videos, TV, we talk, we are planning some field trips Soarin' & Soothin' - Meditation, Relaxation, and Sleep Aid; mental health is important

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