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Find 4 equations of the form A [OPERATION] B = C, using the numbers 1-4 and operations $+,-,\times,\div$. Each number must be used once for the A's, once for the B's, and once for the C's, and each operation must likewise be used once.

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There are two possibilities.

We have the following equations, where each column ($A,B,C$) contains $1,2,3,4$ in some order:

  • $A+B=C$
  • $A-B=C$
  • $A\times B=C$
  • $A\div B=C$

The $B=4$ can't be in the $+$ or $-$ row, so we must have

either $1\times4=4$ or $4\div4=1$.

The $A=4$ can't be in the $+$ row, or (by the above spoilertag) in the $\times$ row; ditto for the $C=1$. So each of these must be either in the $-$ row or the $\div$ row.

The $A=1$ can't be in the $-$ row, or (by the above spoilertag) in the $\div$ row; ditto for the $C=4$. So each of these must be either in the $+$ row or the $\times$ row.


ASSUME we have the first of the two possibilities from the above spoilertag. Then the $+$ row can't have 1, 3, or 4 in the $A$ column, so it must be

$2+1=3$.

The remaining two numbers in the $B$ column are 2 and 3. If $B=2$ is in the $\div$ row, then it must be $4\div2=2$, leaving the only numbers left for the $-$ row as $3-3=1$, contradiction. So our equations are:

  • $2+1=3$

  • $4-2=2$

  • $1\times4=4$

  • $3\div3=1$.


ASSUME we have the second of the two possibilities from the above spoilertag. Then the $-$ row can't have 1, 3, or 4 in the $C$ column, so it must be

$3-1=2$.

The remaining two numbers in the $C$ column are 3 and 4. If $C=3$ is in the $+$ row, then one of $A$ and $B$ in this row must be 2; but also $C=4$ is in the $\times$ row, so both $A$ and $B$ there must be 2; contradiction. So the equations are:

  • $2+2=4$

  • $3-1=2$

  • $1\times3=3$

  • $4\div4=1$

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Got it!

4/4=1
2+2=4
3-1=2
1X3=3

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  • $\begingroup$ Your original version was also (almost) right: you just had to change the 4/2=2 to 4-2=2. See my answer for a complete proof that these two solutions are the only possibilities. $\endgroup$ – Rand al'Thor Jul 11 '17 at 12:15
  • $\begingroup$ @Randal'Thor wow, I feel stupid now, I gave up and found a whole new system when all I had to do is change 1 symbol >.< $\endgroup$ – stack reader Jul 11 '17 at 14:01

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