## A riddle with two/three variables

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
akoj
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### A riddle with two/three variables

My friend told me this riddle:
"A year ago my age was squarable, after one year it will be cubicable. How old am I now?"

So.. by cubing and squaring different numbers you can figure out the current age is 26 (5^2=25 and 3^3=27)
But I can't solve it mathematically

I get stuck at thinking the way to solve it is either:
x^3=2+y^2

or:

x+1 = z^3
x-1 = y^2

Either way, I can't get any further. How do you solve this riddle mathematically? Or is it even possible?

CorkSea
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Joined: Tue Nov 02, 2010 3:39 am
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### Re: A riddle with two/three variables

Need 2 equations:

We know:
y^3 - x^2 = 2

We also know:
x - y = 2

So, we can solve the second equation for x and plug it into the first:
x = 2 + y

Plug into first equation to get:
y^3 - (2 + y)^2 = 2

Now try solving for y ;):
y^3 - (4 + 4y + y^2) = 2
y^3 - 4 - 4y - y^2 = 2
y^3 - y^2 - 4y - 6 = 0

Wolfram alpha tried solving this for me and told me that the real solution to this problem is y = 3. Let's take that and run with it. Plug y = 3 back into your second equation to get:
x - 3 = 2

Solve that for x to get x = 5. We can plug x = 5 and y = 3 back into our first equation to show:
3^3 - 5^2 = 2
27 - 25 = 2

If someone can come up with a 3 variable equation, it might come out nicer (meaning you wouldn't need wolfram alpha to solve it). I'll check back if I think of anything.

akoj
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Joined: Sun Oct 31, 2010 7:50 pm
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### Re: A riddle with two/three variables

If someone can come up with a 3 variable equation, it might come out nicer (meaning you wouldn't need wolfram alpha to solve it). I'll check back if I think of anything.
Thanks, please check back if you figure it out. This is really bugging me, my friend who asked this is a teacher for 8th graders and this riddle is from their book. I'm somewhat older than 8th graders and I can't figure it out.

Another way to solve it is to put x^3=2+y^2 in to a graphical calculator and observe where both y and x are whole digits. Not exactly mathematical either though.

stapel_eliz
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With the information provided, you get two equations in three unknowns. By definition, this cannot be solved algebraically for a unique solution value.

The student is likely expected to note that the solution value must contain only whole-number values, and do some intelligent guess-n-check, probably also using the fact that there aren't many perfect-square and perfect-cube pairs which are reasonable for human ages.