[Guest]

In the following equation, replace x ,y and z with some digits (eg. 1,2,3 etc) and "?" and "!" with some mathematical operator (eg. +-*/), so that the equation hols true

xz ? xy = zx ! xy

This equation has only two known solutions

[Guest]

In the following equation, replace x ,y and z with some digits (eg. 1,2,3 etc) and "?" and "!" with some mathematical operator (eg. +-*/), so that the equation hols true

xz ? xy = zx ! xy

This equation has only two known solutions

Are x, y and z nonzero? If not, then...

01 * 05 = 10 - 05

[Scudo]

X=0

Y=1

Z=2

!=+

?=-

xz ? xy = zx ! xy

[Guest]

X=0

Y=1

Z=2

!=+

?=-

xz ? xy = zx ! xy

That would be 02 - 01 = 20 + 01

[Guest]

There is no stipulation that x,y,z must be unique (or that ! and ? are unique).

So the entire set of cases where x=y=z and !=? are all valid solutions.

15 + 18 = 51 - 18

27 - 24 = 72 / 24

28 + 27 = 82 - 27

49 - 47 = 94 / 47

If you include leading zeros

01 * 05 = 10 - 05

09 + 06 = 90 / 06

02 + 09 = 20 - 09

[Guest]

You can replace all with zeros or all with ones or as kevink said make all the same you'd have a true equation

[Guest]

All ones! Yay one party! (strobe lights and one glyphs dancing)

[Guest]

If we allow x, y and z and/or the arithmetic operations to not be distinct, and/or allow leading zeroes, we have many solutions.

I believe sak_iko meant that x ≠ y, x ≠ z, y ≠ z; x > 0; z > 0; and ? ≠ !.

{(1, 8, 5), (+, -)} :: 15 + 18 = 51 - 18 = 33

{(2, 7, 8), (+, -)} :: 28 + 27 = 82 - 27 = 55

{(2, 4, 7), (-, /)} :: 27 - 24 = 72 / 24 = 3

{(4, 7, 9), (-, /)} :: 49 - 47 = 94 / 47 = 2

Edited July 28, 2010 by Dej Mar

[Guest]

X,Y and Z must be non-zero else there are an infinity of answers and the question is pointless.

[Guest]

X,Y and Z are non-zero and not equal. I have thought about two solutions

27 - 24 = 72 / 24

49 - 47 = 94 / 47

The other two correct solutions are provided by Kevink and Dejmar

28 + 27 = 82 - 27

15 + 18 = 51 - 18

[Guest]

Solved by Kevink and Dej Mar

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