Problem

**DenMac21** · January 30th 2006, 08:40 AM

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
$10x+y=3(x+y)$

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

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

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

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

Is my way of solution right?
help to solve.

**ThePerfectHacker** · January 30th 2006, 09:59 AM

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
$10x+y=3(x+y)$

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

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

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

I have tried to solve this equation by puting $x=\frac{2y}{7}$ (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
$ \frac{10x^2+11xy+y^2}{3}=(x+y)^2 $
You accidently did 2 steps in step 2. You multiplied by $(x+y)$ and divided by 3, then you forgot that you divided by 3 and divided by 3 again.

**DenMac21** · January 30th 2006, 10:55 AM

Originally Posted by ThePerfectHacker

You made a mistake, on step 3, your result should look like
$ \frac{10x^2+11xy+y^2}{3}=(x+y)^2 $
You accidently did 2 steps in step 2. You multiplied by $(x+y)$ 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:
$ 10x+y=3(x+y) $
$ \frac{10x^2+11xy+y^2}{9}=(x+y)^2 $

**CaptainBlack** · January 30th 2006, 08:09 PM

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:
$ 10x+y=3(x+y) $
$ \frac{10x^2+11xy+y^2}{9}=(x+y)^2 $

Since $x \in \{1,2,3,4,5,6,7,8,9\}$ and $y \in \{0,1,2,3,4,5,6,7,8,9\}$ ,
trial and error shows that the first of your equations has solution:

$x=2,\ y=7$ ,

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

RonL

**CaptainBlack** · January 30th 2006, 09:03 PM

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 $Z$ be the number wanted. The sum of the digits of the
two-digit number is $9$ as the digits are $2$ and $7$ . So the last
condition of the problem translates to:

$3Z=9^2=81$ ,

so:

$Z=27$ .

RonL

**DenMac21** · January 31st 2006, 12:44 AM

Originally Posted by CaptainBlack

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

$3Z=9^2=81$ ,

so:

$Z=27$ .

RonL

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

**CaptainBlack** · January 31st 2006, 02:44 AM

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:

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

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

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

RonL

**DenMac21** · January 31st 2006, 03:13 AM

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:

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

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

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

RonL

Thanks, nice solution.

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