Fill in the boxes to get the right equation v2

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
7
down vote

favorite












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



$$Box-Box+BoxtimesBox=Box:/:Box$$



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










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  • 1




    Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
    – Racso
    Sep 22 at 14:06










  • @Rasco Of course it does, how else?
    – Roman Odaisky
    Sep 22 at 14:08














up vote
7
down vote

favorite












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



$$Box-Box+BoxtimesBox=Box:/:Box$$



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










share|improve this question

















  • 1




    Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
    – Racso
    Sep 22 at 14:06










  • @Rasco Of course it does, how else?
    – Roman Odaisky
    Sep 22 at 14:08












up vote
7
down vote

favorite









up vote
7
down vote

favorite











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



$$Box-Box+BoxtimesBox=Box:/:Box$$



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










share|improve this question













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



$$Box-Box+BoxtimesBox=Box:/:Box$$



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







mathematics lateral-thinking no-computers






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share|improve this question











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asked Sep 22 at 14:01









Roman Odaisky

1614




1614







  • 1




    Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
    – Racso
    Sep 22 at 14:06










  • @Rasco Of course it does, how else?
    – Roman Odaisky
    Sep 22 at 14:08












  • 1




    Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
    – Racso
    Sep 22 at 14:06










  • @Rasco Of course it does, how else?
    – Roman Odaisky
    Sep 22 at 14:08







1




1




Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
– Racso
Sep 22 at 14:06




Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
– Racso
Sep 22 at 14:06












@Rasco Of course it does, how else?
– Roman Odaisky
Sep 22 at 14:08




@Rasco Of course it does, how else?
– Roman Odaisky
Sep 22 at 14:08










4 Answers
4






active

oldest

votes

















up vote
4
down vote



accepted










Edit: Got it.




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




Great puzzle!






share|improve this answer






















  • What are you going to do with the genvyvat fynfu?
    – Roman Odaisky
    Sep 22 at 15:38










  • It’s not possible for me to vtaber vg?
    – Excited Raichu
    Sep 22 at 15:45










  • Would be qvivqr ol mreb!
    – Weather Vane
    Sep 22 at 15:47










  • I suppose so. Svkrq vg!
    – Excited Raichu
    Sep 22 at 15:52










  • This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
    – Roman Odaisky
    Sep 22 at 15:56

















up vote
5
down vote













My answer is




$boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $




because




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 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $




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



Solved with pencil and paper only.






share|improve this answer






















  • 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.
    – Roman Odaisky
    Sep 22 at 17:16

















up vote
2
down vote













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).




Proof:




This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ 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, $$Box-Box+BoxtimesBox=1,2;textor;4.$$






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




Proof:




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 $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ 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 $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$







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.




Proof:




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 $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ 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, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$







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





And finally,




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




Proof:




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 $Boxtimes Box$ 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 $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ 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?).




Therefore,




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 $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.







share|improve this answer


















  • 1




    Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
    – PotatoLatte
    Sep 22 at 14:53










  • @PotatoLatte yeah, I know. That's why this answer is partial :P
    – user477343
    Sep 23 at 0:38

















up vote
1
down vote













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).







share|improve this answer






















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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote



    accepted










    Edit: Got it.




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




    Great puzzle!






    share|improve this answer






















    • What are you going to do with the genvyvat fynfu?
      – Roman Odaisky
      Sep 22 at 15:38










    • It’s not possible for me to vtaber vg?
      – Excited Raichu
      Sep 22 at 15:45










    • Would be qvivqr ol mreb!
      – Weather Vane
      Sep 22 at 15:47










    • I suppose so. Svkrq vg!
      – Excited Raichu
      Sep 22 at 15:52










    • This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
      – Roman Odaisky
      Sep 22 at 15:56














    up vote
    4
    down vote



    accepted










    Edit: Got it.




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




    Great puzzle!






    share|improve this answer






















    • What are you going to do with the genvyvat fynfu?
      – Roman Odaisky
      Sep 22 at 15:38










    • It’s not possible for me to vtaber vg?
      – Excited Raichu
      Sep 22 at 15:45










    • Would be qvivqr ol mreb!
      – Weather Vane
      Sep 22 at 15:47










    • I suppose so. Svkrq vg!
      – Excited Raichu
      Sep 22 at 15:52










    • This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
      – Roman Odaisky
      Sep 22 at 15:56












    up vote
    4
    down vote



    accepted







    up vote
    4
    down vote



    accepted






    Edit: Got it.




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




    Great puzzle!






    share|improve this answer














    Edit: Got it.




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




    Great puzzle!







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Sep 22 at 15:51

























    answered Sep 22 at 15:29









    Excited Raichu

    70413




    70413











    • What are you going to do with the genvyvat fynfu?
      – Roman Odaisky
      Sep 22 at 15:38










    • It’s not possible for me to vtaber vg?
      – Excited Raichu
      Sep 22 at 15:45










    • Would be qvivqr ol mreb!
      – Weather Vane
      Sep 22 at 15:47










    • I suppose so. Svkrq vg!
      – Excited Raichu
      Sep 22 at 15:52










    • This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
      – Roman Odaisky
      Sep 22 at 15:56
















    • What are you going to do with the genvyvat fynfu?
      – Roman Odaisky
      Sep 22 at 15:38










    • It’s not possible for me to vtaber vg?
      – Excited Raichu
      Sep 22 at 15:45










    • Would be qvivqr ol mreb!
      – Weather Vane
      Sep 22 at 15:47










    • I suppose so. Svkrq vg!
      – Excited Raichu
      Sep 22 at 15:52










    • This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
      – Roman Odaisky
      Sep 22 at 15:56















    What are you going to do with the genvyvat fynfu?
    – Roman Odaisky
    Sep 22 at 15:38




    What are you going to do with the genvyvat fynfu?
    – Roman Odaisky
    Sep 22 at 15:38












    It’s not possible for me to vtaber vg?
    – Excited Raichu
    Sep 22 at 15:45




    It’s not possible for me to vtaber vg?
    – Excited Raichu
    Sep 22 at 15:45












    Would be qvivqr ol mreb!
    – Weather Vane
    Sep 22 at 15:47




    Would be qvivqr ol mreb!
    – Weather Vane
    Sep 22 at 15:47












    I suppose so. Svkrq vg!
    – Excited Raichu
    Sep 22 at 15:52




    I suppose so. Svkrq vg!
    – Excited Raichu
    Sep 22 at 15:52












    This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
    – Roman Odaisky
    Sep 22 at 15:56




    This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
    – Roman Odaisky
    Sep 22 at 15:56










    up vote
    5
    down vote













    My answer is




    $boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $




    because




    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 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $




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



    Solved with pencil and paper only.






    share|improve this answer






















    • 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.
      – Roman Odaisky
      Sep 22 at 17:16














    up vote
    5
    down vote













    My answer is




    $boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $




    because




    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 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $




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



    Solved with pencil and paper only.






    share|improve this answer






















    • 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.
      – Roman Odaisky
      Sep 22 at 17:16












    up vote
    5
    down vote










    up vote
    5
    down vote









    My answer is




    $boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $




    because




    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 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $




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



    Solved with pencil and paper only.






    share|improve this answer














    My answer is




    $boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $




    because




    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 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $




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



    Solved with pencil and paper only.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Sep 22 at 18:09

























    answered Sep 22 at 16:15









    Weather Vane

    3777




    3777











    • 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.
      – Roman Odaisky
      Sep 22 at 17:16
















    • 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.
      – Roman Odaisky
      Sep 22 at 17:16















    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.
    – Roman Odaisky
    Sep 22 at 17:16




    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.
    – Roman Odaisky
    Sep 22 at 17:16










    up vote
    2
    down vote













    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).




    Proof:




    This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ 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, $$Box-Box+BoxtimesBox=1,2;textor;4.$$






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




    Proof:




    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 $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ 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 $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$







    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.




    Proof:




    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 $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ 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, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$







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





    And finally,




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




    Proof:




    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 $Boxtimes Box$ 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 $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ 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?).




    Therefore,




    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 $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.







    share|improve this answer


















    • 1




      Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
      – PotatoLatte
      Sep 22 at 14:53










    • @PotatoLatte yeah, I know. That's why this answer is partial :P
      – user477343
      Sep 23 at 0:38














    up vote
    2
    down vote













    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).




    Proof:




    This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ 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, $$Box-Box+BoxtimesBox=1,2;textor;4.$$






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




    Proof:




    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 $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ 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 $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$







    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.




    Proof:




    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 $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ 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, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$







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





    And finally,




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




    Proof:




    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 $Boxtimes Box$ 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 $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ 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?).




    Therefore,




    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 $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.







    share|improve this answer


















    • 1




      Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
      – PotatoLatte
      Sep 22 at 14:53










    • @PotatoLatte yeah, I know. That's why this answer is partial :P
      – user477343
      Sep 23 at 0:38












    up vote
    2
    down vote










    up vote
    2
    down vote









    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).




    Proof:




    This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ 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, $$Box-Box+BoxtimesBox=1,2;textor;4.$$






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




    Proof:




    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 $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ 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 $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$







    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.




    Proof:




    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 $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ 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, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$







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





    And finally,




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




    Proof:




    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 $Boxtimes Box$ 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 $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ 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?).




    Therefore,




    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 $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.







    share|improve this answer














    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).




    Proof:




    This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ 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, $$Box-Box+BoxtimesBox=1,2;textor;4.$$






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




    Proof:




    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 $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ 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 $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$







    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.




    Proof:




    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 $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ 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, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$







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





    And finally,




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




    Proof:




    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 $Boxtimes Box$ 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 $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ 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?).




    Therefore,




    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 $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Sep 22 at 15:28

























    answered Sep 22 at 14:37









    user477343

    3,0111748




    3,0111748







    • 1




      Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
      – PotatoLatte
      Sep 22 at 14:53










    • @PotatoLatte yeah, I know. That's why this answer is partial :P
      – user477343
      Sep 23 at 0:38












    • 1




      Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
      – PotatoLatte
      Sep 22 at 14:53










    • @PotatoLatte yeah, I know. That's why this answer is partial :P
      – user477343
      Sep 23 at 0:38







    1




    1




    Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
    – PotatoLatte
    Sep 22 at 14:53




    Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
    – PotatoLatte
    Sep 22 at 14:53












    @PotatoLatte yeah, I know. That's why this answer is partial :P
    – user477343
    Sep 23 at 0:38




    @PotatoLatte yeah, I know. That's why this answer is partial :P
    – user477343
    Sep 23 at 0:38










    up vote
    1
    down vote













    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).







    share|improve this answer


























      up vote
      1
      down vote













      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).







      share|improve this answer
























        up vote
        1
        down vote










        up vote
        1
        down vote









        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).







        share|improve this answer














        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).








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Sep 22 at 15:42

























        answered Sep 22 at 14:27









        Racso

        4906




        4906



























             

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