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How should I approach using two 8s and two 3s to make the number 24?


Coppers - Make 24$tau$ is greatest! $tau$ is all (from 1 to 20)Largest number with five 1's and five numeric operations90s Number PuzzleHow many consecutive integers can you make using only four digits?Use four 8's to make the number 2016Fix the perfectly ordered equationUse 2, 0, 1 and 8 to make 199Palindromic number puzzle - make 505 from 20202Use 0, 5, 7 and 1 to make 89Use 6, 5 and 3 to make 57













9












$begingroup$



Use two $8$s, two $3$s, and basic arithmetic operators ($ +, -, times , div$, parentheses) to make the number $24$.



(You may not join numbers together to form new numbers, like $ 8, 3rightarrow 83$)




I don't know how to start besides just trying to find the correct answer. Is there a way you can make this equation through small steps or I should just bruteforce it?










share|improve this question









New contributor




user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$












  • $begingroup$
    see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
    $endgroup$
    – JonMark Perry
    17 hours ago






  • 1




    $begingroup$
    Possible duplicate of Coppers - Make 24
    $endgroup$
    – North
    13 hours ago






  • 8




    $begingroup$
    If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
    $endgroup$
    – Lunin
    13 hours ago










  • $begingroup$
    One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with if no, and opens up more options if yes.
    $endgroup$
    – Justin Time
    13 hours ago
















9












$begingroup$



Use two $8$s, two $3$s, and basic arithmetic operators ($ +, -, times , div$, parentheses) to make the number $24$.



(You may not join numbers together to form new numbers, like $ 8, 3rightarrow 83$)




I don't know how to start besides just trying to find the correct answer. Is there a way you can make this equation through small steps or I should just bruteforce it?










share|improve this question









New contributor




user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
    $endgroup$
    – JonMark Perry
    17 hours ago






  • 1




    $begingroup$
    Possible duplicate of Coppers - Make 24
    $endgroup$
    – North
    13 hours ago






  • 8




    $begingroup$
    If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
    $endgroup$
    – Lunin
    13 hours ago










  • $begingroup$
    One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with if no, and opens up more options if yes.
    $endgroup$
    – Justin Time
    13 hours ago














9












9








9


1



$begingroup$



Use two $8$s, two $3$s, and basic arithmetic operators ($ +, -, times , div$, parentheses) to make the number $24$.



(You may not join numbers together to form new numbers, like $ 8, 3rightarrow 83$)




I don't know how to start besides just trying to find the correct answer. Is there a way you can make this equation through small steps or I should just bruteforce it?










share|improve this question









New contributor




user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





Use two $8$s, two $3$s, and basic arithmetic operators ($ +, -, times , div$, parentheses) to make the number $24$.



(You may not join numbers together to form new numbers, like $ 8, 3rightarrow 83$)




I don't know how to start besides just trying to find the correct answer. Is there a way you can make this equation through small steps or I should just bruteforce it?







mathematics formation-of-numbers






share|improve this question









New contributor




user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 21 mins ago









Omega Krypton

4,5751440




4,5751440






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Check out our Code of Conduct.









asked 20 hours ago









user9888273user9888273

563




563




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New contributor





user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user9888273 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
    $endgroup$
    – JonMark Perry
    17 hours ago






  • 1




    $begingroup$
    Possible duplicate of Coppers - Make 24
    $endgroup$
    – North
    13 hours ago






  • 8




    $begingroup$
    If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
    $endgroup$
    – Lunin
    13 hours ago










  • $begingroup$
    One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with if no, and opens up more options if yes.
    $endgroup$
    – Justin Time
    13 hours ago


















  • $begingroup$
    see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
    $endgroup$
    – JonMark Perry
    17 hours ago






  • 1




    $begingroup$
    Possible duplicate of Coppers - Make 24
    $endgroup$
    – North
    13 hours ago






  • 8




    $begingroup$
    If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
    $endgroup$
    – Lunin
    13 hours ago










  • $begingroup$
    One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with if no, and opens up more options if yes.
    $endgroup$
    – Justin Time
    13 hours ago
















$begingroup$
see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
$endgroup$
– JonMark Perry
17 hours ago




$begingroup$
see puzzling.stackexchange.com/questions/50259/coppers-make-24 (GM's answer)
$endgroup$
– JonMark Perry
17 hours ago




1




1




$begingroup$
Possible duplicate of Coppers - Make 24
$endgroup$
– North
13 hours ago




$begingroup$
Possible duplicate of Coppers - Make 24
$endgroup$
– North
13 hours ago




8




8




$begingroup$
If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
$endgroup$
– Lunin
13 hours ago




$begingroup$
If I am not mistaken, this question is not asking people to solve the puzzle in question, but is instead asking strategies for how to go about solving it beyond just trying things at random.
$endgroup$
– Lunin
13 hours ago












$begingroup$
One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with if no, and opens up more options if yes.
$endgroup$
– Justin Time
13 hours ago




$begingroup$
One thing I'd suggest is determining whether any rounding is allowed. Narrows down the number of pieces you have to work with if no, and opens up more options if yes.
$endgroup$
– Justin Time
13 hours ago










5 Answers
5






active

oldest

votes


















23












$begingroup$

While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)



Here's one method of thinking to get to the answer:



1) Is this a trick question?



It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.



2) What do we need to do?



What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.



3) Can we simplify the problem down at all?



Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.



4) What components get us closer to the solution?



So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?





  • $1 = 8 div 8$ with $3,3$ leftover


  • $16 = 8 + 8$ with $3,3$ leftover

  • $64 = 8 times 8$

  • $0 = 8 - 8$


  • $24 = 8 times 3$ with $8,3$ leftover

  • $11 = 8 + 3$

  • $5 = 8 - 3$

  • $2 frac{2}{3} = 8 div 3$

  • ...


5) Work from the other end - what do the components of the solution look like?



Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:




$sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$




I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?




  • $8 * 3 = 24$

  • $8 / frac{1}{3} = 24$

  • $27 - 3 = 24$

  • $21 + 3 = 24$

  • $32 - 8 = 24$

  • ...


6) Connect the dots.



We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?



From the above, here's the link I've come up with:




$ 3 - 2 frac{2}{3} = frac{1}{3}$




That will lead us to a solution by putting it all together:




$8 div (3 - frac{8}{3})) = 24$




Fin



That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.






share|improve this answer











$endgroup$









  • 13




    $begingroup$
    Hey look, someone actually answered the question asked! :)
    $endgroup$
    – Rubio
    10 hours ago



















12












$begingroup$

Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).




$8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
$= 8 div frac{1}{3}$
$= 24$




If we allow square roots, a simpler solution is possible.




$sqrt{8 times 8 times 3 times 3}$
$= 8 times 3$
$= 24$




In fact, there are many more solutions if you allow more operations.






share|improve this answer











$endgroup$





















    4












    $begingroup$

    Here's a fairly simple one:




    $3!times(frac88+3)$







    share|improve this answer









    $endgroup$









    • 4




      $begingroup$
      Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
      $endgroup$
      – eedrah
      10 hours ago



















    0












    $begingroup$

    Another simple solution :




    $(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$







    share|improve this answer










    New contributor




    Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$









    • 1




      $begingroup$
      Factorials are not allowed, unfortunately....
      $endgroup$
      – Glorfindel
      8 hours ago



















    -1












    $begingroup$

    Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:




    $(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$







    share|improve this answer









    $endgroup$













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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      23












      $begingroup$

      While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)



      Here's one method of thinking to get to the answer:



      1) Is this a trick question?



      It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.



      2) What do we need to do?



      What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.



      3) Can we simplify the problem down at all?



      Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.



      4) What components get us closer to the solution?



      So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?





      • $1 = 8 div 8$ with $3,3$ leftover


      • $16 = 8 + 8$ with $3,3$ leftover

      • $64 = 8 times 8$

      • $0 = 8 - 8$


      • $24 = 8 times 3$ with $8,3$ leftover

      • $11 = 8 + 3$

      • $5 = 8 - 3$

      • $2 frac{2}{3} = 8 div 3$

      • ...


      5) Work from the other end - what do the components of the solution look like?



      Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:




      $sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$




      I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?




      • $8 * 3 = 24$

      • $8 / frac{1}{3} = 24$

      • $27 - 3 = 24$

      • $21 + 3 = 24$

      • $32 - 8 = 24$

      • ...


      6) Connect the dots.



      We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?



      From the above, here's the link I've come up with:




      $ 3 - 2 frac{2}{3} = frac{1}{3}$




      That will lead us to a solution by putting it all together:




      $8 div (3 - frac{8}{3})) = 24$




      Fin



      That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.






      share|improve this answer











      $endgroup$









      • 13




        $begingroup$
        Hey look, someone actually answered the question asked! :)
        $endgroup$
        – Rubio
        10 hours ago
















      23












      $begingroup$

      While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)



      Here's one method of thinking to get to the answer:



      1) Is this a trick question?



      It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.



      2) What do we need to do?



      What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.



      3) Can we simplify the problem down at all?



      Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.



      4) What components get us closer to the solution?



      So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?





      • $1 = 8 div 8$ with $3,3$ leftover


      • $16 = 8 + 8$ with $3,3$ leftover

      • $64 = 8 times 8$

      • $0 = 8 - 8$


      • $24 = 8 times 3$ with $8,3$ leftover

      • $11 = 8 + 3$

      • $5 = 8 - 3$

      • $2 frac{2}{3} = 8 div 3$

      • ...


      5) Work from the other end - what do the components of the solution look like?



      Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:




      $sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$




      I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?




      • $8 * 3 = 24$

      • $8 / frac{1}{3} = 24$

      • $27 - 3 = 24$

      • $21 + 3 = 24$

      • $32 - 8 = 24$

      • ...


      6) Connect the dots.



      We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?



      From the above, here's the link I've come up with:




      $ 3 - 2 frac{2}{3} = frac{1}{3}$




      That will lead us to a solution by putting it all together:




      $8 div (3 - frac{8}{3})) = 24$




      Fin



      That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.






      share|improve this answer











      $endgroup$









      • 13




        $begingroup$
        Hey look, someone actually answered the question asked! :)
        $endgroup$
        – Rubio
        10 hours ago














      23












      23








      23





      $begingroup$

      While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)



      Here's one method of thinking to get to the answer:



      1) Is this a trick question?



      It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.



      2) What do we need to do?



      What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.



      3) Can we simplify the problem down at all?



      Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.



      4) What components get us closer to the solution?



      So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?





      • $1 = 8 div 8$ with $3,3$ leftover


      • $16 = 8 + 8$ with $3,3$ leftover

      • $64 = 8 times 8$

      • $0 = 8 - 8$


      • $24 = 8 times 3$ with $8,3$ leftover

      • $11 = 8 + 3$

      • $5 = 8 - 3$

      • $2 frac{2}{3} = 8 div 3$

      • ...


      5) Work from the other end - what do the components of the solution look like?



      Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:




      $sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$




      I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?




      • $8 * 3 = 24$

      • $8 / frac{1}{3} = 24$

      • $27 - 3 = 24$

      • $21 + 3 = 24$

      • $32 - 8 = 24$

      • ...


      6) Connect the dots.



      We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?



      From the above, here's the link I've come up with:




      $ 3 - 2 frac{2}{3} = frac{1}{3}$




      That will lead us to a solution by putting it all together:




      $8 div (3 - frac{8}{3})) = 24$




      Fin



      That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.






      share|improve this answer











      $endgroup$



      While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)



      Here's one method of thinking to get to the answer:



      1) Is this a trick question?



      It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.



      2) What do we need to do?



      What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.



      3) Can we simplify the problem down at all?



      Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.



      4) What components get us closer to the solution?



      So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?





      • $1 = 8 div 8$ with $3,3$ leftover


      • $16 = 8 + 8$ with $3,3$ leftover

      • $64 = 8 times 8$

      • $0 = 8 - 8$


      • $24 = 8 times 3$ with $8,3$ leftover

      • $11 = 8 + 3$

      • $5 = 8 - 3$

      • $2 frac{2}{3} = 8 div 3$

      • ...


      5) Work from the other end - what do the components of the solution look like?



      Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:




      $sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$




      I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?




      • $8 * 3 = 24$

      • $8 / frac{1}{3} = 24$

      • $27 - 3 = 24$

      • $21 + 3 = 24$

      • $32 - 8 = 24$

      • ...


      6) Connect the dots.



      We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?



      From the above, here's the link I've come up with:




      $ 3 - 2 frac{2}{3} = frac{1}{3}$




      That will lead us to a solution by putting it all together:




      $8 div (3 - frac{8}{3})) = 24$




      Fin



      That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 8 hours ago









      boboquack

      15.5k149118




      15.5k149118










      answered 10 hours ago









      eedraheedrah

      875118




      875118








      • 13




        $begingroup$
        Hey look, someone actually answered the question asked! :)
        $endgroup$
        – Rubio
        10 hours ago














      • 13




        $begingroup$
        Hey look, someone actually answered the question asked! :)
        $endgroup$
        – Rubio
        10 hours ago








      13




      13




      $begingroup$
      Hey look, someone actually answered the question asked! :)
      $endgroup$
      – Rubio
      10 hours ago




      $begingroup$
      Hey look, someone actually answered the question asked! :)
      $endgroup$
      – Rubio
      10 hours ago











      12












      $begingroup$

      Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).




      $8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
      $= 8 div frac{1}{3}$
      $= 24$




      If we allow square roots, a simpler solution is possible.




      $sqrt{8 times 8 times 3 times 3}$
      $= 8 times 3$
      $= 24$




      In fact, there are many more solutions if you allow more operations.






      share|improve this answer











      $endgroup$


















        12












        $begingroup$

        Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).




        $8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
        $= 8 div frac{1}{3}$
        $= 24$




        If we allow square roots, a simpler solution is possible.




        $sqrt{8 times 8 times 3 times 3}$
        $= 8 times 3$
        $= 24$




        In fact, there are many more solutions if you allow more operations.






        share|improve this answer











        $endgroup$
















          12












          12








          12





          $begingroup$

          Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).




          $8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
          $= 8 div frac{1}{3}$
          $= 24$




          If we allow square roots, a simpler solution is possible.




          $sqrt{8 times 8 times 3 times 3}$
          $= 8 times 3$
          $= 24$




          In fact, there are many more solutions if you allow more operations.






          share|improve this answer











          $endgroup$



          Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).




          $8 div (3 - (8 div 3))$ (or alternatively $frac{8}{3 - frac{8}{3}}$)
          $= 8 div frac{1}{3}$
          $= 24$




          If we allow square roots, a simpler solution is possible.




          $sqrt{8 times 8 times 3 times 3}$
          $= 8 times 3$
          $= 24$




          In fact, there are many more solutions if you allow more operations.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 19 hours ago

























          answered 20 hours ago









          HughHugh

          2,2651926




          2,2651926























              4












              $begingroup$

              Here's a fairly simple one:




              $3!times(frac88+3)$







              share|improve this answer









              $endgroup$









              • 4




                $begingroup$
                Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
                $endgroup$
                – eedrah
                10 hours ago
















              4












              $begingroup$

              Here's a fairly simple one:




              $3!times(frac88+3)$







              share|improve this answer









              $endgroup$









              • 4




                $begingroup$
                Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
                $endgroup$
                – eedrah
                10 hours ago














              4












              4








              4





              $begingroup$

              Here's a fairly simple one:




              $3!times(frac88+3)$







              share|improve this answer









              $endgroup$



              Here's a fairly simple one:




              $3!times(frac88+3)$








              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 18 hours ago









              JonMark PerryJonMark Perry

              19.5k63991




              19.5k63991








              • 4




                $begingroup$
                Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
                $endgroup$
                – eedrah
                10 hours ago














              • 4




                $begingroup$
                Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
                $endgroup$
                – eedrah
                10 hours ago








              4




              4




              $begingroup$
              Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
              $endgroup$
              – eedrah
              10 hours ago




              $begingroup$
              Not sure if I'd call factorial 'basic', but I love the ingenuity here. It's even inspired me for crazy solution
              $endgroup$
              – eedrah
              10 hours ago











              0












              $begingroup$

              Another simple solution :




              $(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$







              share|improve this answer










              New contributor




              Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$









              • 1




                $begingroup$
                Factorials are not allowed, unfortunately....
                $endgroup$
                – Glorfindel
                8 hours ago
















              0












              $begingroup$

              Another simple solution :




              $(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$







              share|improve this answer










              New contributor




              Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$









              • 1




                $begingroup$
                Factorials are not allowed, unfortunately....
                $endgroup$
                – Glorfindel
                8 hours ago














              0












              0








              0





              $begingroup$

              Another simple solution :




              $(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$







              share|improve this answer










              New contributor




              Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$



              Another simple solution :




              $(8-3)!/(8-3) = 5!/5 = (5*4*3*2*1)/5 = 4*3*2*1 = 24$








              share|improve this answer










              New contributor




              Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.









              share|improve this answer



              share|improve this answer








              edited 7 hours ago









              Omega Krypton

              4,5751440




              4,5751440






              New contributor




              Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.









              answered 8 hours ago









              Srinivas TamtamSrinivas Tamtam

              9




              9




              New contributor




              Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.





              New contributor





              Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              Srinivas Tamtam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.








              • 1




                $begingroup$
                Factorials are not allowed, unfortunately....
                $endgroup$
                – Glorfindel
                8 hours ago














              • 1




                $begingroup$
                Factorials are not allowed, unfortunately....
                $endgroup$
                – Glorfindel
                8 hours ago








              1




              1




              $begingroup$
              Factorials are not allowed, unfortunately....
              $endgroup$
              – Glorfindel
              8 hours ago




              $begingroup$
              Factorials are not allowed, unfortunately....
              $endgroup$
              – Glorfindel
              8 hours ago











              -1












              $begingroup$

              Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:




              $(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$







              share|improve this answer









              $endgroup$


















                -1












                $begingroup$

                Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:




                $(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$







                share|improve this answer









                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:




                  $(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$







                  share|improve this answer









                  $endgroup$



                  Inspired from JonMark Perry's solution, here's another crazy 'non basic' one:




                  $(( 8 times 8 ) % 3 + 3)! = (64 % 3 + 3)! = (1 + 3)! = 24$








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 10 hours ago









                  eedraheedrah

                  875118




                  875118






















                      user9888273 is a new contributor. Be nice, and check out our Code of Conduct.










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                      user9888273 is a new contributor. Be nice, and check out our Code of Conduct.













                      user9888273 is a new contributor. Be nice, and check out our Code of Conduct.












                      user9888273 is a new contributor. Be nice, and check out our Code of Conduct.
















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