Age Mystery

Number Theory Level 3

An old man was playing with his grandson before the boy started asking about their ages.

Boy : Grandpa, what will happen to our ages 7 years from now?
Grandpa : The sum of our ages will be a perfect square.

Boy : Oh, and then what will happen 7 years after that?
Grandpa : Then the ratio of my age to yours will be a perfect square.

Boy : Really? At that time will our ages combine to become 100?
Grandpa : No, my boy, but if we add your present age to the sum of our ages in 14 years, then it will be 100.

What is the sum of their present ages?

The answer is 67.

1 solution

Dan Ley
Nov 27, 2016

Let the boy's age be $x$ and the Grandpa's age be $y$ , where $x<y$ . From the text, we can immediately infer three equations:

$x+y+14=a^2$ (where $a$ is a positive integer)
$\frac{y+14}{x+14}=b^2$ (where $b$ is also a positive integer)
$2x+y+28=100 \implies 2x+y=72$

Since $x < y$ , the final equation tells us that $3x< 2x + y = 72$ and hence that $x < 24$ .

From the final equation, we can rearrange for $x+y=72-x$ and substitute this into the first equation, yielding $86-x=a^2$ . This means that $x=5,22,37,50,61$ .

Since $x<24$ , we need only to consider $x=5$ , $y=62$ and $x=22$ , $y=28$ .

Case 1: $x = 22, y = 28$
Checking the second equation, we get $\frac{ 28 + 14} { 22 + 14} = \frac{ 42 }{36}$ which is not a perfect square

Case 2: $x = 5, y = 62$
Checking the second equation, we get that $\frac{ 62 + 14 } { 5 + 14 } = \frac{ 76}{19} = 4$ which is a perfect sqaure.

Therefore the boy's age is $5$ , the Grandpa's age is $62$ and their combined age is $67$ . These values can be verified by plugging them back into the three equations.

Nice solution by bounding these values.

Note: The problem doesn't require that " $b$ is a positive integer distinct from $a$ ". Yes, that is ultimately a true statement, but it is not a stated condition.

Calvin Lin Staff - 4 years, 6 months ago

You're right that I was wrong to assume that " $b$ is a positive integer distinct from $a$ " without justification. I've edited the post, but I can now prove that $a\neq b$ using a proof by contradiction.

Let's assume that $a=b$ . We have that $y+14=a^2-x$ (by rearranging the first equation). Applying this to the second equation gives: $\frac{a^2-x}{x+14}=a^2\implies a^2-x=a^2x+14a^2 \implies -13a^2=x(a^2+1)$ .

Tidying this shows us that $\frac{-13a^2}{a^2+1}=x$ , implying that $x$ is always negative. This we know to be false, so the assumption that $a=b$ must be wrong, and we have that " $b$ is a positive integer distinct from $a$ ".

Dan Ley - 4 years, 6 months ago

My point is that you subsequently do not use that fact in the rest of the solution, so bringing it up would unnecessarily confuse the reader / lead them down an irrelevant path.

Calvin Lin Staff - 4 years, 6 months ago

@Calvin Lin – True, I'll try to be more succinct next time.

Dan Ley - 4 years, 6 months ago

@Worranat Pakornrat what is meant by that sum is it after 7 years or after more 7 years , you are making obscure references

Syed Hissaan - 4 years, 3 months ago

14 years from present if you follow the boy's conversation.

Worranat Pakornrat - 4 years, 3 months ago

in the conversation , you have used the word sum in the first part (i.e after 7 years ) it would be be better if you mention No, my boy, but if we add your present age to that sum( after 14 years), it will be 100. , so that the other solvers don't have to face difficulty

Syed Hissaan - 4 years, 3 months ago

@Syed Hissaan – Thanks. I've added that in for clarity :)

Calvin Lin Staff - 4 years, 3 months ago

Age Mystery

The answer is 67.

1 solution

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