Math riddle

[MXracer16]

A man is trying to figure out the ages of 3 kids while taking a census. One lady is very mad at him and wants to give him a hard time about figuring out the ages because he interupted her shower.He is givin the following information.

The product of their ages is 36.

The sum of their ages is the house number.

All of the kids are younger then the child next door.

What are the ages of the 3 kids?

[ZSK]

There is not enough information given that would result in a single answer. A few possibilites are:

1, 1, 36
1, 2, 18
1, 3, 12
1, 4, 9
1, 6, 6
2, 2, 9
2, 3, 6
3, 3, 4

You don't know the age of the women, the house number, or the age of any individual child, there could be twins, and you don't know the age of the child next door.

[MXracer16]

I dont know, thats all the information we were givin. I dont know if this would help but he said the lady answered the door with 9 towels on.

[ZSK]

9 towels wouldn't change the age of her kids.;) Is this 5th grade math?

[MXracer16]

Im stumped, this is extra credit in my Geometry class. Usually he gives us these kind of problems and you cant think too hard about them. One of the clues usually has some stupid way of answering the problem.

[ZSK]

Any and all of the answers I listed would be correct with the information you supplied. He could always ask the kids or the father the ages of the childeren.

[MXracer16]

I dunno, Ill see what the answer is tomorrow.

[TGW_400ex]

Is it 4?

[rooster_20]

You need to know the house number.

[Scro]

I googled it and came up with this answer -



The special properties of 36 are VERY important here. You need to see
that the complete factorization of 36 is 2*2*3*3. You need this to get
the most out of these clues. Let's use A, B, and C for the 3 ages,
and assume they are in order from younger to older. They could all be
different like A = 2, B = 3, C = 6 or some could be the same like
A = 3, B = 3, C = 4 (she could have 3 year old twins). Now make a
table of ALL the possibilities. Don't forget the ones with the first
age being one (1). Also make a column for the sum of the childrens'
ages:

A B C sum
--- --- --- -----
1 1 36 38
1 6 6 13
2 3 6 11
3 3 4 10

The table continues on. I got a total of 8 possibilities in my table.
Remember A <= B <= C. Now, my first row, with a grown child 36 years
old and 1-year-old twins is unlikely, but we have to consider all the
possibilities so we don't miss anything. OK?

Now on to clue No. 2. For each line in the table, you should add up
the ages and enter that value in the right-hand most column. You will
get a different sum of ages A + B + C for most of the rows, BUT two of
the rows will come out with the same sum. That must be the house
number across the street, because otherwise the second clue would be
enough to show which set of ages was right. Before you go on to clue
No. 3, be sure you have all eight rows of numbers in the table
completely filled out, and that you see which 2 rows have the same sum
in the last column.

Now on to clue No. 3. This third clue could just as well be "the
oldest one has 12 toes". The real clue here is the word "oldest" which
eliminates the possibility of one younger child and two older twins.
That's because the phrase "the oldest one" implies only one. If you
did all the work along with me, you know the answer.

[parkers30]

Originally posted by rooster_20
You need to know the house number.

yep, by the 'child next door' I guess you could assume one would not be a legal adult (ie. 18 or 36) but that is stretching it

Originally posted by Scro
I googled it and came up with this answer -



The special properties of 36 are VERY important here. You need to see
that the complete factorization of 36 is 2*2*3*3. You need this to get
the most out of these clues. Let's use A, B, and C for the 3 ages,
and assume they are in order from younger to older. They could all be
different like A = 2, B = 3, C = 6 or some could be the same like
A = 3, B = 3, C = 4 (she could have 3 year old twins). Now make a
table of ALL the possibilities. Don't forget the ones with the first
age being one (1). Also make a column for the sum of the childrens'
ages:

A B C sum
--- --- --- -----
1 1 36 38
1 6 6 13
2 3 6 11
3 3 4 10

The table continues on. I got a total of 8 possibilities in my table.
Remember A <= B <= C. Now, my first row, with a grown child 36 years
old and 1-year-old twins is unlikely, but we have to consider all the
possibilities so we don't miss anything. OK?

Now on to clue No. 2. For each line in the table, you should add up
the ages and enter that value in the right-hand most column. You will
get a different sum of ages A + B + C for most of the rows, BUT two of
the rows will come out with the same sum. That must be the house
number across the street, because otherwise the second clue would be
enough to show which set of ages was right. Before you go on to clue
No. 3, be sure you have all eight rows of numbers in the table
completely filled out, and that you see which 2 rows have the same sum
in the last column.

Now on to clue No. 3. This third clue could just as well be "the
oldest one has 12 toes". The real clue here is the word "oldest" which
eliminates the possibility of one younger child and two older twins.
That's because the phrase "the oldest one" implies only one. If you
did all the work along with me, you know the answer.



f'in cheater :blah:

[parkers30]

this has a different clue 3 though, if you by the same logic, we are down to :

1,6,6
2,2,9

due to the sums being the same at 13

[Scro]

Originally posted by mjpridered10
f'in cheater :blah:

Atleast I can admit it;) :D

[ZSK]

Originally posted by Scro
I googled it and came up with this answer....If you
did all the work along with me, you know the answer.

That doesn't even address that the house number isn't given. House numbers can range from five to a single digit. So "the sum of their ages is the house number" is irrelevant without a supplied house number even the house next door would suffice.

All of the kids are younger then the child next door. The child next door may be 86 with a 100 year old mother. This statement is of no significance to the problem without the age of a child that lives next door.

The only thing with any value to the problem (assuming that you've given all of the information in the problem) is that the product of their ages is 36, to which the answers have been given.

[Scro]

You can't think of the house number as a number that you are supposed to know, but rather that the woman knows her own house number. It doesn't matter what the number is.

[ZSK]

How can you meet that requirement of the answer if you don't know the house number? Obviously the women knows her own house number as does the man recording the census information.

[ZSK]

Originally posted by Scro
I googled it and came up with this answer -

Now on to clue No. 2. For each line in the table, you should add up
the ages and enter that value in the right-hand most column. You will
get a different sum of ages A + B + C for most of the rows, BUT two of
the rows will come out with the same sum. That must be the house
number across the street, because otherwise the second clue would be
enough to show which set of ages was right. Before you go on to clue

How is the house across the street going to be the same as the house in question? Houses are numbered even on one side of the street and odd on the other.

[Scro]

Since she didn't know the ages after the second clue, it must mean that her street address appears twice - which does appear twice as 13. Otherwise, she would have known what the answer was without even looking at the third clue.

What's throwing it off, is the fact that it is not worded correctly.

[ZSK]

What's throwing it off is that it is wrong. It would be the same as putting two gallons of water in a one gallon jug and getting three back out of your one gallon jug.

[parkers30]

he right in a sense Zack, on somewhat of a techinicality. At the same time it is dead WRONG because there is no reason one would have to use the third clue.....

there just isn't enough info given

[ilpadrino113]

either that was more information that you missed or your teacher missed becasue there isn't enough info there. the only part that would matter would be to find out what 3 numbers are products of 36, and any of those combos could be it. The hosue number could be 5318 so any age would be less, and the "kid" next door could be 40.

[Chino]

12, 12, 12

[ilpadrino113]

12, 12, 12


ummm..... no


the product of 3 12's = 1728


but close tho:rolleyes:

[ZSK]

Originally posted by parkers30
he right in a sense Zack, on somewhat of a techinicality. At the same time it is dea WRONG because there is no reason one would have to use the third clue.....

there just isn't enough info given

I agree that the theory is correct, at least to the first "clue." The sum would need to equal the missing house number but there is significance that two sum of two of the possibilties are equal it only shows that there could be two answers that meet the requirements.

I don't see how it's right even on a technicality. How could a house be numbered 13 while the house across the street is also numbered 13? It would seem logical for that house to be 12 or 14.

It comes down to there just isn't enough information given, like you said.

[MXracer16]

Its 2, 2, and 9. I did word it wrong in my first post. Stupid me. :( Its supposed to say the oldest is younger then then the next door neighbor. I feel like an idiot, I didnt realize I copied the problem wrong until I called someone.

To figure it out this is how it goes. You make the table you guys all did, then you add up the ages you got for each part of the table. You will get 2 answers that are the same; 2, 2, and 9, and 1, 6, and 6. But it said that there is an oldest one, so that eliminates 1, 6, and 6, so therefore it has to be 2, 2, and 9.

[Chino]

Originally posted by ilpadrino113
ummm..... no


the product of 3 12's = 1728


but close tho:rolleyes:

my bad I read it as saying sum, not product, chill my man ~

[wilkin250r]

Who in the world said that there was another house with the number 13?

There is obviously a reason for the 3rd clue. If the house number were anything else, the third clue wouldn't be needed, AND we would need the house number to solve the problem (or else a different third clue).

But since the third clue has nothing to do with house numbers, we have to assume the house number is 13, because it's the only one with two solutions.

Never anywhere does it say that there is another house with the same number. Just because you have two possible solutions with the number 13 doesn't mean there are two different houses with the number 13.

[MXracer16]

Haha, turns out I had the problem right from the beginning. The person I called copied the problem wrong. Try and figure it out using my first post. It IS possible, so dont tell me there isnt enough information. Have fun. :devil:

[rooster_20]

Im still sticking with my answer of there isnt enough information given. In all equations there is only a possible answer if there are one unknown. In this equation the possibilities are varied because there are more than one unknowns. EXAMPLE: 2+x=5 you know that x=3. This equation is more like x+x=5 and there could be multiple answers.

[MXracer16]

Originally posted by rooster_20
Im still sticking with my answer of there isnt enough information given. In all equations there is only a possible answer if there are one unknown. In this equation the possibilities are varied because there are more than one unknowns. EXAMPLE: 2+x=5 you know that x=3. This equation is more like x+x=5 and there could be multiple answers.

Nope, there is a way to get one, and only one answer.

[Scro]

Alright, I think we give up:p