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I am currently working on a little extra credit for my 9th grade math class and I am stuck on the one of the problems.

The Three-Number Problem

I have chosen three numbers. The second is twice the first, and the third is three times the second. The sum of the first two when multiplied by the sum of the last two happens to be the same as the first number multiplied by the quare of the second number.

I decided to approach this by first starting with 3 variables.

f, s, and t.

f standing for the first number.
s standing for the second number.
t standing for the third number.

f = #
s = 2f
t = 3s

The equation I came up with was:

(f + s) * (s + t) = fs2

Which can be simplified to:

fs + ft + s2 = fs2

However, from there on, I am stuck. I have no idea how to solve for f. However, i'm sure that i'm in the right place and that somebody here can help me out. Thank you!

Community
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Shawn Holman
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    In the last equation, instead of writing $t$, write $3s$, and then instead of writing $s$, write $2f$. – Andrés E. Caicedo Oct 07 '13 at 00:52
  • Andres's suggestion is good, and to have thought of it yourself, notice that your simplified equation doesn't take into account what you know about which numbers are multiples of the others. His suggested substitution incorporates everything you know into one place. Since you mentioned solving for $f$, I'll note that you could solve your simplified equation for $f$ right away, though you'll eventually have to do what Andres suggests. There are no powers of $f$, so you can collect all terms with $f$ on one side and factor out an $f$. – Steve Kass Oct 07 '13 at 01:05
  • Your key mistake was using three variables. Just use one! You have enough information to only use one. You can't solve that last equation because it's got three variables. – Jack M Oct 07 '13 at 07:36
  • Another way of seeing there's a problem is that that last equation does not contain the information that $s=2f$ and $t=3s$. – Jack M Oct 07 '13 at 07:42

2 Answers2

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So you have $f$; $s = 2f$; and $t = 3s = 6f$.

Furthermore, $(f+s)(s+t) = fs^2$.

Rewriting the latter equation with only $f$, this would mean:

$$(f+2f)(2f + 6f) = f(4f^2)$$

Can you now solve for $f$?

Since the other answer gives full details:

Either $f = 0$ and we are done, or divide both sides by $f^2$ to obtain $24 = 4f$.

Then $f = 6$. QED

Benjamin Dickman
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Welcome to Math.SE.

Your approach is quite correct so far, though you made a mistake. You wrote that:

$(f + s) \cdot (s + t) = fs^{2}$

Which can be simplified to:

$fs + ft + s^{2} = fs^{2}$

Whereas it really simplifies to: $fs + ft + s^{2} + st = fs^{2}$.

Let's use your definition of the variables to do some substitution.

$fs + ft + s^{2} + st = fs^{2}$

Substitute $t$ for $3s$.

$fs + f3s + s^{2} + s(3s) = fs^{2}$.

$fs + f3s + s^{2} + 3s^{2} = fs^{2}$.

$fs + f3s + 4s^{2} = fs^{2}$.

Substitute $s$ for $2f$.

$f(2f) + f(6f) + 4(2f)^{2} = f(2f)^{2}$

$2f^{2} + 6f^{2} + 4(4f^{2}) = f(4f^{2})$

$2f^{2} + 6f^{2} + 16f^{2} = f(4f^{2})$

$24f^{2} = 4f^{3}$.

$6f^{2} = f^{3}$

Divide both sides by $f^{2}$.

$6 = f$

Newb
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