My take on a recent puzzle. Place numbers into boxes to create a correct equation, using each number at least once:


This time the four decimal numbers are 3, 6, 7, 12.

  • 1
    $\begingroup$ Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it? $\endgroup$ – Racso Sep 22 '18 at 14:06
  • $\begingroup$ @Rasco Of course it does, how else? $\endgroup$ – Roman Odaisky Sep 22 '18 at 14:08

Edit: Got it.

-$7 + 3 * 3 = 12/6$ (Still don’t have to fill in every box.)

Great puzzle!

  • $\begingroup$ What are you going to do with the genvyvat fynfu? $\endgroup$ – Roman Odaisky Sep 22 '18 at 15:38
  • $\begingroup$ It’s not possible for me to vtaber vg? $\endgroup$ – Excited Raichu Sep 22 '18 at 15:45
  • $\begingroup$ Would be qvivqr ol mreb! $\endgroup$ – Weather Vane Sep 22 '18 at 15:47
  • $\begingroup$ I suppose so. Svkrq vg! $\endgroup$ – Excited Raichu Sep 22 '18 at 15:52
  • $\begingroup$ This is indeed the intended solution. I wonder whether someone can out-lateral me though :−) $\endgroup$ – Roman Odaisky Sep 22 '18 at 15:56

My answer is

$\boxed{12} - \boxed{12} + \boxed{\: 3 \: } \times \boxed{\: 3 \: } = \boxed{63} \: / \: \boxed{\: 7 \: }$


the question says "place numbers into boxes" so I placed two numbers into the 5th box.

Note: There is one solution to the obvious question:

$\boxed{\: 7 \: } - \boxed{12} + \boxed{\: 3 \: } \times \boxed{\: 3 \: } = \boxed{12} \: / \: \boxed{\: 3 \: }$

But it is illegal because it does not use all the numbers.

Solved with pencil and paper only.

  • $\begingroup$ Also very nice. I was pretty sure repeated application of this idea can lead to a solution, but to solve the puzzle doing it only once looks remarkable to me. $\endgroup$ – Roman Odaisky Sep 22 '18 at 17:16

Partial Answer:

Well, well, well.

THERE EXISTS NO SOLUTION! (without lateral thinking, although inverting the $6$ to a $9$ has no solution either, in particular).

Let's call the numbers we can choose from, the Option Numbers.

The right hand side (RHS) will either equal $1$, $2$ or $4$ (without lateral thinking).


This is our equation: $$\Box-\Box+\Box\times\Box=\Box\:/\:\Box\tag{$\small\rm given$}$$ $7$ can only be in the fifth box if the sixth box is also $7$, as that is the only number out of the option numbers that divide $7$. Otherwise, the left hand side (LHS) would not be an integer. So, a possibility is that the RHS is equal to $7/7=1$.

Excluding $7$ now, $3$ cannot be in the fifth box because it is the lowest option number, and thus the fraction will not be an integer otherwise. That leaves $6$ and $12$.

So the fraction is either $12/6$, $12/3$ or $6/3$ which is $2$, $2$ or $4$ respectively. Since $2=2$, then the RHS is either equal to $2$ or $4$. But we mentioned that it can also be equal to $1$, so the possible values the RHS can equal are $$1,2,4.$$

Therefore, $$\Box-\Box+\Box\times\Box=1,2\;\text{or}\;4.$$

$\Box-\Box$ equals a number in between $-14$ and $-83$ inclusive.


From the previous proof, we know that the LHS is either $1$, $2$ or $4$. All the option numbers are positive, and the minimum product that can be made from them is $3\times 6$ or $6\times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $\Box-\Box$ is... $$\Box-\Box=4-18=-14.\tag{$\small\rm as \; the \; maximum \; value$}$$ Now, doing the opposite to find the minimum value, we find maximum product of option numbers and minimum value RHS can equal. That makes $84$ and $1$ respectively (since $84=12\times 7$ or $7\times 12$). So the minimum value of $\Box-\Box$ is... $$\Box-\Box=1-84=-83.\tag{$\small\rm as \; the \; minimum \; value$}$$

Therefore, the expression, $\Box-\Box$, must equal something in between $-14$ and $-83$ inclusive. Since it is negative, the first box must have an option number smaller than the second box. Now we know where to place numbers, we can make this range between $-14$ and $-83$ smaller!

$\Box-\Box$ must equal something in between $-1$ and $-9$ inclusive.


Let the first box be the lowest option number (namely, $3$) and the second box be the highest option number (namely, $12$). We obtain that $$\boxed{3}-\boxed{12}=-9.\tag{$\small\rm as \; the \; minimum \; value$}$$ This is the minimum value because the pair $(3,12)$ is the furthest away from each other out of all the option numbers; and since we are dealing with negatives, the maximum (furthest away) turns to minimum (because positive becomes negative). Now look at the option numbers: $3$, $6$, $7$, $12$. Which pair is the closest to each other? That pair is $(6,7)$ which differs by only $1$. Therefore, $$\boxed{6}-\boxed{7}=-1.\tag{$\small\rm as \; the \; maximum \; value$}$$

And therefore, $\Box-\Box$ must equal something in between $-1$ and $-9$ inclusive.

And finally,

The maximum value of $\Box\times \Box$ must be $12$... uh oh. That doesn't seem right... it's a contradiction; THERE EXISTS NO SOLUTION (without lateral thinking)!


If we get the maximum value the LHS can equal (namely, $4$) and subtract the minimum value of $\Box-\Box$ (namely, $-9$), we get $$4-(-9)=4+9=13.$$ Therefore, the maximum value of $\Box\times \Box$ is $13$. But $13$ is prime, so the maximum value reduces to $12$.

But we now have a problem. The minimum product that can be made from the option numbers is $3\times 6$ or $6\times 3$ which is $18$. Therefore, the minimum value of $\Box\times \Box$ is $18$. $18>12$. This is a contradiction. We do not need to consider the minimum value of LHS anymore, now (can you guess why?).


THERE IS NO SOLUTION! (without lateral thinking).

This was tougher than the puzzle that inspired the OP... but I did it. In fact,

Invert $6$ to make $9$ and substitute in all the proofs (with minimum value of product being $7\times 3$ or $3\times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.

  • 1
    $\begingroup$ Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P $\endgroup$ – PotatoLatte Sep 22 '18 at 14:53
  • $\begingroup$ @PotatoLatte yeah, I know. That's why this answer is partial :P $\endgroup$ – Mr Pie Sep 23 '18 at 0:38

OK, so:

$$ 12-7+6*3 = 2E_{16}/10_{2} $$ You restrict the decimal numbers we can use to 3, 6, 7 and 12, but you don't restrict other numeric bases. Here, I'm using a base 16 number (2E, equivalent to 46 in base 10) and a binary number (10, equivalent to 2).


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