You made a mistake, on step 3, your result should look likeOriginally Posted byDenMac21

You accidently did 2 steps in step 2. You multiplied by and divided by 3, then you forgot that you divided by 3 and divided by 3 again.

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- Jan 30th 2006, 08:40 AM #1

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## 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

Multiplying both sides with (x+y) we get that

Dividing left side with 3 we will get that

which is number that we want.

So I think we must solve system of equation

I have tried to solve this equation by puting (from first equation) into second equation but the result is x=0 and y=0.

Is my way of solution right?

help to solve.

- Jan 30th 2006, 09:59 AM #2

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- Jan 30th 2006, 10:55 AM #3

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Originally Posted by**ThePerfectHacker**

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:

- Jan 30th 2006, 08:09 PM #4

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- Jan 30th 2006, 09:03 PM #5

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- Jan 31st 2006, 12:44 AM #6

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- Jan 31st 2006, 02:44 AM #7

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Originally Posted by**DenMac21**

And if you don't like exaustive search you can try the following:

,

but and are prime and so is a multiple of ,

and as is the only single digit non-zero multiple of

we have and the rest follows.

( has to be non-zero, as if it were zero would be zero,

which is not permitted).

RonL

- Jan 31st 2006, 03:13 AM #8

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