There is something wrong with the equation below:

enter image description here

By just exchanging two squares at a time, find the equality with the least amount of exchange.

For example if this question is asked for below equation, you may find the solution with one time exchange:

enter image description here

the answer will be exchanging "+" and "9" which makes the equation correct as below:


FYI: Even though it is not a hint, all numbers are used only once. ($0,1,2,3,4,5,6,7,8,9$)

Hint: you only swap each square once for optimal clear solution.

  • 1
    $\begingroup$ Do the squares we swap have to be adjacent like in the example? $\endgroup$ Commented Sep 5, 2018 at 13:56
  • $\begingroup$ @JaapScherphuis no, you can swap any square with any other square you want $\endgroup$
    – Oray
    Commented Sep 5, 2018 at 14:05
  • 1
    $\begingroup$ A few questions – (1) can I place the brackets beside a number (like $5(6)$) to do multiplication, (2) are we doing left-to-right or order of operations? $\endgroup$
    – user46002
    Commented Sep 5, 2018 at 14:23
  • 1
    $\begingroup$ @Hugh left to right and you may yes $\endgroup$
    – Oray
    Commented Sep 5, 2018 at 14:36

3 Answers 3


@oray's edit: Here is my intented answer since no-one found it out:

$3140\times(6+9)=2-75\times8$ (Initial layout)
$3140\times(6+9)-2=75\times8$ ("$=$" and "$-$")
$3\times40\times(6+9)-2=7518$ ("$\times$" and "$1$")
$3\times45\times(6+9)-2=7018$ ("$0$" and "$5$")
$3\times45\times(6+9)-7=2018$ ("$2$" and "$7$")

which is

the year we are in.

I found another solution with

4 exchanges

which is:

$3140\times(6+9)=2-75\times8$ (Initial layout)
$3+40\times(619)=2-75\times8$ (Swapped $1$ and $+$)
$3+40=(619)\times2-75\times8$ (Swapped $=$ and $\times$)
$3+40=(629)\times1-75\times8$ (Swapped $2$ and $1$)
$5+40=(629)\times1-73\times8$ (Swapped $5$ and $3$)
To arrive at $45=45$


Original post: I'm not sure if this is optimal, but I can do it in

4 exchanges

like this:

$3140\times(6+9)=2-75\times8$ (Initial layout)
$314\times0(6+9)=2-75\times8$ (Swapped 0 and $\times$)
$314\times0(6+9)=27-5\times8$ (Swapped 7 and $-$)
$314\times0(6+8)=27-5\times9$ (Swapped 8 and 9)
$514\times0(6+8)=27-3\times9$ (Swapped 3 and 5)
To arrive at $0=0$.

  • $\begingroup$ this is good one, but not my intended answer and 4 is actually optimal though actual answer does not require any trick. $\endgroup$
    – Oray
    Commented Sep 5, 2018 at 21:15
  • $\begingroup$ @Oray what do you mean by "no trick"? this answer fits all the rules, does it not? $\endgroup$
    – Quintec
    Commented Sep 6, 2018 at 2:02
  • $\begingroup$ @Quintec originally i was not planning to accept 0( concept then i thought it would be okay :) $\endgroup$
    – Oray
    Commented Sep 6, 2018 at 5:21
  • $\begingroup$ According to a comment on the OP, evaluation should be left to right, which would give $27-3\times9=216$. FWIW, I'd prefer usual operator precedence in any arithmetic puzzle. $\endgroup$
    – nickgard
    Commented Sep 6, 2018 at 8:06
  • $\begingroup$ Hopefully the new solution I found was your intended answer @Oray :) $\endgroup$ Commented Sep 6, 2018 at 8:54

I don't think it's optimal but I can do it in

$6$ exchanges.


Label the boxes, from left to right, by $A,B,C,\ldots,Q$. Then the exchanges are:

(i) $J \leftrightarrow M$
$3140\times(6+9 -=2)75\times8$

(ii) $L \leftrightarrow M$
$3140\times(6+9 -=)275\times8$

(iii) $K \leftrightarrow M$
$3140\times(6+9 -2)=75\times8$

(iv) $I \leftrightarrow K$
$3140\times(6+2 -9)=75\times8$

(v) $A \leftrightarrow I$
$2140\times(6+3 -9)=75\times8$

(vi) $D \leftrightarrow Q$
$2148\times(6+3 -9)=75\times0$


Well, to start it off, I can do

9 exchanges(thanks to @Jaap Scherphuis for reading the rules)

to end with


the details of which I have not quite figured out how to display in the answer.

  • $\begingroup$ Since non-adjacent exchanges are allowed, the above answer can actually be done in 9. It is the permutation (CJHIGFE)(LQ)(MOP) where the squares are labelled A to Q, which needs no more than 6+1+2=9 exchanges. $\endgroup$ Commented Sep 5, 2018 at 14:51
  • $\begingroup$ @JaapScherphuis thanks, my bad, didn't read the rules carefully $\endgroup$
    – Quintec
    Commented Sep 5, 2018 at 17:34

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