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March 30, 2015

March 30, 2015

Posted by **Devin** on Monday, February 14, 2011 at 11:19am.

+ ACB

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BCA

- Math -
**tchrwill**, Monday, February 14, 2011 at 12:35pmWhat three different digits are represented by X,Y,Z in this addition problem?

XZY + XYZ = YZX

Just another perspective:

1--First, from the last column, we could have either Y + Z = X or Y + Z = 10 + X

2--From the second column, we could have either Y + Z = Z or Y + Z + 1 = 10 + Z

3--But we can't have (Y + Z) equaling both X and Z, so Y and Z must add up to more than 10.

3--Now we have Y + Z = 10 + X and Y + Z + 1 = 10 + Z or Y + Z = Z + 9.

4--Equating (Y + Z)'s, we get 10 + X = Z + 9 or X = Z - 1

5--Knowing that Y + Z exceeds 10, we can say 2X + 1 = Y

6--Substituting, we get 2X + 1 + Z = 10 + X or X = 9 - Z

7--Equating our expressions for X we get 9 - Z = Z - 1 or 2Z = 10 and Z = 5.

I thing the rest should fall out fairly easy for you.

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