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Correct Digits For Correct Letters
The equation in question was....
First of all X + Y + Z = Z < = 10 in not possible since in that case, X + Y + Z - Z =0 i.e.
X + Y = 0.
Hence, there must be carry 1 forwarded to digit's place. So,
X + Y + Z = 10 + Z
X + Y = 10. ........(1)
Therefore, possible pairs for (X,Y) or for (Y,X) are (1,9), (2,8), (3,7), (4,6) and (5,5) out of which (5,5) is invalid as repeating digits are not allowed.
Now, since carry is forwarded to ten's place addition, it looks like,
1 + X + Y + Z = XY
Putting (1) in above,
1 + 10 + Z = XY
11 + Z = XY ........(2)
The maximum value of Z can be 9 & in that case as per above equation XY = 20. But since 0 is not allowed Z must be less than or equal to 8. Then XY <= 11 + 8 = 19.
So X which seems to be carry to hundred place must be 1 (can't be 0 or 2 as proved above). Hence, X = 1.
If X = 1, then from (1), Y = 9.
And if XY = 19 then from (2) Z = 8.
To conclude, X = 1, Y = 9 and Z = 8.
Final equation, looks like,
11 + 99 + 88 = 198.
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Another Method :
The given equation is nothing but
10X + X + 10Y + Y + 10Z + Z = 100X + 10Y + Z
11(X + Y + Z) = 100X + 10Y + Z
89X = Y + 10Z = 10Z + Y
Since, only whole digits are allowed, maximum value of 10Z + Y can be 98 with Z = 9 & Y = 8. So if X = 2 then 89 x 2 = 198 = 10Z + Y is impossible case. Hence, X = 1.
And only Z = 8 and Y = 9 satisfies 89 x 1 = 10Z + Y = 10(8) + 9 = 89.
In short, X = 1, Y = 9 and Z = 8.
Flip The Orientation of the Group
Assume these matchsticks as fishes. Can you flip the positions of the fishes horizontally i.e. 1 at the leftmost position and 4 at the rightmost position. Condition is that you can move only 3 matchsticks.
That's how it can be done!
That's how it can be done!
