# Thread: algebra puzzle

1. ## algebra puzzle

I have one problem that I can't seem to solve or even get started on. Here's the problem:

The letters A-I stand for the numbers 1-9. Use the clues to find the number each letter represents. Each number will only be used once.

A · A =GA
G · B =A
I · B =H
A / D= A
F / G = G
F – C =B

A=
B=
C=
D=
E=
F=
G=
H=
I=

Explain in words how you eliminated certain numbers for specific letters. Also explain how you figured out what number any of the letters must represent.

Any hints, tips, would be greatly appreciated. Thank you.

2. ## Re: algebra puzzle

Originally Posted by woodyl
I have one problem that I can't seem to solve or even get started on. Here's the problem:

The letters A-I stand for the numbers 1-9. Use the clues to find the number each letter represents. Each number will only be used once.

A · A =GA
G · B =A
I · B =H
A / D= A
F / G = G
F – C =B

A=
B=
C=
D=
E=
F=
G=
H=
I=

Explain in words how you eliminated certain numbers for specific letters. Also explain how you figured out what number any of the letters must represent.

Any hints, tips, would be greatly appreciated. Thank you.

A · A =GA
Does this mean : A multiplied by A it is equal to G multiplied by A ??

3. ## Re: algebra puzzle

$\displaystyle A\cdot{A}=GA$
$\displaystyle 3<A<10$ as $\displaystyle 3\cdot{3}=09$, which doesn't fit.

Test square numbers:
$\displaystyle 4^2=16$, so $\displaystyle A$ can't be $\displaystyle 4$.
$\displaystyle 5^2=25$, so $\displaystyle A$ could be $\displaystyle 5$.
$\displaystyle 6^2=36$, so $\displaystyle A$ could be $\displaystyle 6$
$\displaystyle 7^2=49$ - nope.
$\displaystyle 8^2=64$ - nope.
$\displaystyle 9^2=81$ - nope.

So $\displaystyle A$ is $\displaystyle 5$, and $\displaystyle G$ is $\displaystyle 2$
OR $\displaystyle A$ is $\displaystyle 6$, and $\displaystyle G$ is $\displaystyle 3$

Knowing, however, that $\displaystyle G\cdot B=A$ is enough to tell us which is the case. From there, you have enough to continue.

4. ## Re: algebra puzzle

Hello, woodyl!

The letters $\displaystyle A$-$\displaystyle I$ stand for the digits $\displaystyle 1$-$\displaystyle 9.$
Use the clues to find the number each letter represents.
Each number will only be used once.

$\displaystyle [1]\;A\cdot A \,=\,GA$

$\displaystyle [2]\;G \cdot B\,=\,A$

$\displaystyle [3]\;I\cdot B \,=\,H$

$\displaystyle [4]\;\frac{A}{D}\,=\,A \quad\Rightarrow\quad A\cdot D \,=\,A$

$\displaystyle [5]\;\frac{F}{G} \,=\, G \quad\Rightarrow\quad G\cdot G \,=\,F$

$\displaystyle [6]\;F – C \,=\,B \quad\Rightarrow\quad F \,=\,B+C$

From [4], we see that: $\displaystyle \boxed{D = 1}$

[1] says that $\displaystyle A^2$ is a two-digit number ending in $\displaystyle A.$
There are only two choices: .$\displaystyle 5^2 \,=\,25\text{ and }6^2 \,=\,36$

If $\displaystyle A = 5$, then $\displaystyle G = 2.$
. . Then [2] becomes: .$\displaystyle 2\cdot B \,=\,5$ . . . impossible.

Hence: .$\displaystyle \boxed{A = 6}\;\boxed{G = 3}$

Then [2] becomes: .$\displaystyle 3\cdot B \,=\,6 \quad\Rightarrow\qyad \boxed{B \,=\,2}$

And [5] becomes: .$\displaystyle 3\cdot3 \,=\,F \quad\rightarrow\quad \boxed{F \,=\,9}$

And [6] becomes: .$\displaystyle 9 \,=\,2 + C \quad\Rightarrow\quad \boxed{C \,=\,7}$

[3] becomes: .$\displaystyle I \cdot 2 \,=\,H$
The only choices are: .$\displaystyle \boxed{I\,=\,4}\;\boxed{H \,=\,8}$

Finally, by elimination: .$\displaystyle \boxed{E \,=\,5}$

. . .$\displaystyle \text{Solution}$

. . $\displaystyle \begin{array}{ccc}A&=&6 \\ B&=&2 \\ C&=&7 \\ D&=&1 \\ E&=&5 \\ F&=&9 \\ G&=&3 \\ H&=&8 \\ I&=&4 \end{array}$

5. ## Re: algebra puzzle

Thank you all for the reponses. I wasn't even close to thinking of the problem that way. I learned to think another way with this. I'm glad our teacher suggested your site.

Thanks again.