1. ## Problem

I need help to solve this problem:

Two-digit number is three times bigger from sum of digits which that number contains, and square of that sum of digits is equal to triple number which is wanted. Find that number.

I have represented that two-digit number as
$\displaystyle 10x+y=3(x+y)$

Multiplying both sides with (x+y) we get that
$\displaystyle \frac{10x^2+11xy+y^2}{3}=(x+y)^2$

Dividing left side with 3 we will get that
$\displaystyle \frac{10x^2+11xy+y^2}{9}=(x+y)^2$
which is number that we want.

So I think we must solve system of equation
$\displaystyle 10x+y=3(x+y)$
$\displaystyle \frac{10x^2+11xy+y^2}{9}=(x+y)^2$

I have tried to solve this equation by puting $\displaystyle x=\frac{2y}{7}$ (from first equation) into second equation but the result is x=0 and y=0.

Is my way of solution right?
help to solve.

2. Originally Posted by DenMac21
I need help to solve this problem:

Two-digit number is three times bigger from sum of digits which that number contains, and square of that sum of digits is equal to triple number which is wanted. Find that number.

I have represented that two-digit number as
$\displaystyle 10x+y=3(x+y)$

Multiplying both sides with (x+y) we get that
$\displaystyle \frac{10x^2+11xy+y^2}{3}=(x+y)^2$

Dividing left side with 3 we will get that
$\displaystyle \frac{10x^2+11xy+y^2}{9}=(x+y)^2$
which is number that we want.

So I think we must solve system of equation
$\displaystyle 10x+y=3(x+y)$
$\displaystyle \frac{10x^2+11xy+y^2}{9}=(x+y)^2$

I have tried to solve this equation by puting $\displaystyle x=\frac{2y}{7}$ (from first equation) into second equation but the result is x=0 and y=0.

Is my way of solution right?
help to solve.
You made a mistake, on step 3, your result should look like
$\displaystyle \frac{10x^2+11xy+y^2}{3}=(x+y)^2$
You accidently did 2 steps in step 2. You multiplied by $\displaystyle (x+y)$ and divided by 3, then you forgot that you divided by 3 and divided by 3 again.

3. Originally Posted by ThePerfectHacker
You made a mistake, on step 3, your result should look like
$\displaystyle \frac{10x^2+11xy+y^2}{3}=(x+y)^2$
You accidently did 2 steps in step 2. You multiplied by $\displaystyle (x+y)$ and divided by 3, then you forgot that you divided by 3 and divided by 3 again.

No, I divided by 3 in step 3 because the square of sum of digits is equal to triple number which is wanted. So I didn't have to divide it by 3 but later it should be divided.

The problem is that I can't find values of x and y.

I can't solve this system of equation:
$\displaystyle 10x+y=3(x+y)$
$\displaystyle \frac{10x^2+11xy+y^2}{9}=(x+y)^2$

4. Originally Posted by DenMac21
No, I divided by 3 in step 3 because the square of sum of digits is equal to triple number which is wanted. So I didn't have to divide it by 3 but later it should be divided.

The problem is that I can't find values of x and y.

I can't solve this system of equation:
$\displaystyle 10x+y=3(x+y)$
$\displaystyle \frac{10x^2+11xy+y^2}{9}=(x+y)^2$
Since $\displaystyle x \in \{1,2,3,4,5,6,7,8,9\}$ and $\displaystyle y \in \{0,1,2,3,4,5,6,7,8,9\}$,
trial and error shows that the first of your equations has solution:

$\displaystyle x=2,\ y=7$,

but this is not a solution of the second of your equations.

RonL

5. Originally Posted by DenMac21
I need help to solve this problem:

Two-digit number is three times bigger from sum of digits which that number contains, and square of that sum of digits is equal to triple number which is wanted. Find that number.
Let $\displaystyle Z$ be the number wanted. The sum of the digits of the
two-digit number is $\displaystyle 9$ as the digits are $\displaystyle 2$ and $\displaystyle 7$. So the last
condition of the problem translates to:

$\displaystyle 3Z=9^2=81$,

so:

$\displaystyle Z=27$.

RonL

6. Originally Posted by CaptainBlack
Let $\displaystyle Z$ be the number wanted. The sum of the digits of the
two-digit number is $\displaystyle 9$ as the digits are $\displaystyle 2$ and $\displaystyle 7$. So the last
condition of the problem translates to:

$\displaystyle 3Z=9^2=81$,

so:

$\displaystyle Z=27$.

RonL
Thats ok, but we have get that x=2 and y=7 by guessing numbers.
Is there algebraic way to get that numbers?

7. Originally Posted by DenMac21
Thats ok, but we have get that x=2 and y=7 by guessing numbers.
Is there algebraic way to get that numbers?
Not by guessing, exhausive search over the ranges involved is not guessing.
And if you don't like exaustive search you can try the following:

$\displaystyle 10x+y=3(x+y) \Rightarrow 7x=2y$,

but $\displaystyle 7$ and $\displaystyle 2$ are prime and so $\displaystyle y$ is a multiple of $\displaystyle 7$,
and as $\displaystyle 7$ is the only single digit non-zero multiple of $\displaystyle 7$
we have $\displaystyle y=7$ and the rest follows.

($\displaystyle y$ has to be non-zero, as if it were zero $\displaystyle x$ would be zero,
which is not permitted).

RonL

8. Originally Posted by CaptainBlack
Not by guessing, exhausive over the ranges involved is not guessing.
And if you don't like exaustive search you can try the following:

$\displaystyle 10x+y=3(x+y) \Rightarrow 7x=2y$,

but $\displaystyle 7$ and $\displaystyle 2$ are prime and so $\displaystyle y$ is a multiple of $\displaystyle 7$,
and as $\displaystyle 7$ is the only single digit non-zero multiple of $\displaystyle 7$
we have $\displaystyle y=7$ and the rest follows.

($\displaystyle y$ has to be non-zero, as if it were zero $\displaystyle x$ would be zero,
which is not permitted).

RonL
Thanks, nice solution.