# Help me to solve math puzzle

### Help me to solve math puzzle

Here is the problem.

A man had nine children, all born at regular intervals, and the sum of the squares of their ages was equal to the square of his own. What was the age of each? Every age was an exact number of years.

burgess

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### Re: Help me to solve math puzzle

Let
$$p$$ be the age of the parent
$$m$$ be the age of the middle child
$$d$$ the difference in ages between successive children

The question states $$p^2 = (m-4d)^2+(m-3d)^2+(m-2d)^2+(m-d)^2+m^2+(m+d)^2+(m+2d)^2+(m+3d)^2+(m+4d)^2$$
which expands and cancels to
$$p^2 = 9m^2 + 60d^2$$
Clearly the right hand side is a multiple of $$3$$ which implies $$p^2$$ is a multiple of $$3$$ and therefore $$p$$ is a multiple of $$3$$. Let $$p = 3t$$, so now we have
$$9t^2= 9m^2 + 60d^2$$
which rearranges to
$$(t+m)(t-m) = 20d^2/3$$
The left hand side is an integer which implies $$d^2$$ and therefore $$d$$ is a multiple of $$3$$.

In order for the youngest child to be born we must have $$m-4d\geq 0$$ which implies $$m\geq 4d$$ and so
$$p^2\geq 9(4d)^2+60d^2 = 204d^2 \geq (14d)^2$$
$$p \geq 14d$$
We know $$d$$ is a multiple of $$3$$, so $$d$$ must be $$0, 3 ,$$ or $$6$$, any higher and $$p$$ would be at least $$126$$ years old which is unreasonable.

Case 1: d = 0
There are lots of solutions if we allow the children to be nonuplets for example $$p=30, m=10,$$ however I don't think there is a documented case of nonuplets surviving for more than a year after birth (however octuplets are known to have survived). This is an unlikely solution.

Case 2: d = 3
From our previous work we know
$$(t+m)(t-m) = 60$$
We also know that $$p = 3t>m+12$$ (parent is older than the oldest child) and $$m\geq 12$$ (youngest child is at least 0), this means that
$$3t>24$$ which implies $$t>8$$ and $$t+m>20$$
Since $$t+m$$ and $$t-m$$ are factors of $$60$$ that leaves us with two possibilities
$$t+m = 60, t-m = 1$$ (which fails as this implies $$t = 30.5$$)
or
$$t+m = 30, t-m = 2$$ which implies $$t = 16, m= 14$$.
So the age of the parent $$p = 48$$, the youngest child is $$2$$, the oldest is $$26$$ and the interval is $$3$$ years.

Case 3: d = 6
We proceed similarly to the previous case.
From our previous work we know
$$(t+m)(t-m) = 240$$
We also know that $$p = 3t>m+24$$ (parent is older than the oldest child) and $$m\geq 24$$ (youngest child is at least 0), this means that
$$3t>48$$ which implies $$t>16$$ and $$t+m>40$$
Since $$t+m$$ and $$t-m$$ are factors of $$240$$ that leaves us with several possibilities:
$$(t+m,t-m) =$$
$$(48,5)$$ [fails as $$t$$ is not an integer]
$$(60,4)$$
$$(80,3)$$ [fails as $$t$$ is not an integer]
$$(120,2)$$ [fails as $$t=61$$ implies the parent's age is $$183$$]
$$(240,1)$$ [fails as $$t$$ is not an integer]
This leaves us with
$$t+m = 60, t-m = 4$$ which implies $$t = 32, m= 28$$.
So the age of the parent $$p = 96,$$ the youngest child is $$4$$, the oldest is $$52$$ and the interval is $$6$$ years. This is an unlikely solution but theoretically possible.

Hope this helped,

R. Baber.
Guest

### Re: Help me to solve math puzzle

Oh the solution is too lengthy.

jackwilson

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