Brain Teasers
Too Many Children
A man has nine children born at regular intervals. The sum of the square of their ages is equal to the square of his own age. What are the ages of his children?
Answer
The ages are 2,5,8,11,14,17,20,23 and 26. The man's age is 48.Hide Answer Show Answer
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Comments
Could you show the math on that?
no clue how you came up with that
I didn't even know where to begin! Good one!
good job, all u havta do is find the square of their ages, add 'em together, then u have his age. After that u just havta figure out that the interval is 3. I did not figure this out, but after reading the answer, i easily figured out how to do it. Good job!
Can you prove that that's the only solution?
Why the low popularity rating? I guess some people do not like math.
explain explain! dun leave us blank n curious on how to arrive at that solution...
Very well put, I liked this teaser.
Please explain. I understand the answer, but how do you come up with the answer? I had too many variables when i tried to set up this problem. (A+B+C+D+E+F+G+H+I)^2 = J^2 Also, the first step is to take the square root of both sides, so why do they need to be squared? I think it would be easier to find if you said that the children had three years in between them and you gave us the age of the children.
I liked the challenge to my brain even though I never figured out the answer. After I looked at the answer, I still never figured out how you arrived at it. I would love to see a formula for how you set this one up! It's a devil of a teaser!
May 19, 2005
I agree with brianz. There is more than one solution. I still had fun figuring that out. Here's the formula:
A=AGE OF THE YOUNGEST CHILD
I=INTERVAL OF AGES
D=DAD'S AGE
A^2+(A+I)^2+(A+2I)^2...=D^2
(continues to the ninth child increasing the interval factor by 1)
All that simplifies to:
9A^2+72AI+204I^2=D^2
A=AGE OF THE YOUNGEST CHILD
I=INTERVAL OF AGES
D=DAD'S AGE
A^2+(A+I)^2+(A+2I)^2...=D^2
(continues to the ninth child increasing the interval factor by 1)
All that simplifies to:
9A^2+72AI+204I^2=D^2
PPPPPlease tell us how u got it
nooooooooo 48^2 is 2304, and 26^2 is not 2304 therefore it does not work.
May 25, 2005
There is no closed form answer to this question.
The answer is a classic hunt and peck searching for the square root of a sum that happens to be a round number. This is because there is only one equation involving 3 variables, the man's age at the birth of first child, the interval between births. and the interval between the last birth and the current age.
The answer is a classic hunt and peck searching for the square root of a sum that happens to be a round number. This is because there is only one equation involving 3 variables, the man's age at the birth of first child, the interval between births. and the interval between the last birth and the current age.
Robo06--
I got the same formula as you did.
What puzzles me is that the normal form for the square of a binomial is:
(a+b)^2=a^2+2ab+b^2.
In this case, the formula is not in
that form. That is, in the polynomial,
9a^2+72ab+204b^2, the sum of the
square roots of the outer terms
doubled is NOT equal to the middle
term.
Can anyone explain?
Jim
I got the same formula as you did.
What puzzles me is that the normal form for the square of a binomial is:
(a+b)^2=a^2+2ab+b^2.
In this case, the formula is not in
that form. That is, in the polynomial,
9a^2+72ab+204b^2, the sum of the
square roots of the outer terms
doubled is NOT equal to the middle
term.
Can anyone explain?
Jim
this question is not worded correctly. in order for that answer to be true, the question must state that all the numbers are whole numbers. otherwise, there is an infinate amount of solutions.
I'm sorry to tell all you guys, but that was way to easy to be called a hard teaser. I used paper and pencil and got the answer in about 3 minutes only. maybe you should just have paid attention to your math teachers.
Hey elvenber,
Since I am one who must have slept through math class, could you answer my question posted above?
We would all appreciate it.
Since I am one who must have slept through math class, could you answer my question posted above?
We would all appreciate it.
Good teaser. totally stumped.
Libra...the idea is 48^2 = 26^2+23^2+20^2 etc etc. Thats why it says the sum of
Libra...the idea is 48^2 = 26^2+23^2+20^2 etc etc. Thats why it says the sum of
what the heck. thats all im gonna say
The equation that many of you got stuck with is called a Diophantine equation (after Diophantus the 3rd century mathematician). A diophantine equation is an equation with more than one variable (obviously) whose solutions must be integers. These equations cannot be solved so easily. You have to eliminate certain cases, and through trial and error and alot of deduction, you come upon an answer which satisfies the said requirments. For those who are about ready to start pulling their hair out, don't despair. After all, Fermat's last theorem was a diophantine equation, and it wasn't solved (or rather proven unsolvable) for about 350 years!!!
Cheers!
Cheers!
Jun 07, 2005
There is a second solution:
Children are 4, 10, 16, 22, 28, 34, 40, 46 and 52.
The man's age is 96.
Not as likely a scenario as the posted answer, but certainly within the realm of possibility.
As far as I can tell there are no other solutions with whole numbers and the man's age under 100.
Children are 4, 10, 16, 22, 28, 34, 40, 46 and 52.
The man's age is 96.
Not as likely a scenario as the posted answer, but certainly within the realm of possibility.
As far as I can tell there are no other solutions with whole numbers and the man's age under 100.
Jun 07, 2005
There are also oddball solutions, such as:
3 sets of triplets, ages 11, 11, 11, 23, 23, 23, 35, 35, 35. Man is age 75.
Or, the degnerative:
a set of nontuplets (???), all age x, with the man age 3x. (e.g. 15 and 45)
Or, triplets aged 7,7,7, 25, 25, 25, 43, 43, 43 and the man age 87.
You could argue that 3 sets of triplets born at equal intervals meets the criteria.
3 sets of triplets, ages 11, 11, 11, 23, 23, 23, 35, 35, 35. Man is age 75.
Or, the degnerative:
a set of nontuplets (???), all age x, with the man age 3x. (e.g. 15 and 45)
Or, triplets aged 7,7,7, 25, 25, 25, 43, 43, 43 and the man age 87.
You could argue that 3 sets of triplets born at equal intervals meets the criteria.
Those are possiblilities only if you disregard the statment that the children were born at regular intervals. But there is only one solution which makes sense for this problem as it is stated, with the ages being integers.
Actually I stand corrected. Your first alternate solution is just as viable. By multiplying both sides by a factor of 2, the solution still makes sense. But it would not work for a factor of three and more (in the real world). Good job!
how are you supposed to get this when you don't know how old any of the kids are??????/ I don't get it at all
I too came up with Robo06's equation. It really comes down to trial and error, which is fine, but I definitely prefer having a way to use some math to at least reduce the possibilities to a reasonable volume.
another one of those teasers that dont actually have a real solution. more like a trivia question.
i like my math teasers to have a solution using math or a little logic, not relying solely on trial and error.
i like my math teasers to have a solution using math or a little logic, not relying solely on trial and error.
Yeah, while my heart is with Pating's comment, it was pretty cool. I wouldn't have seen it in trick, so I'm glad it was here.
A scipe 5-liner ( https://pastebin.com/H4BxuSW6 ) shows that any multiple of 48 is a possible (but not so probable) solution.
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