A Simple Math Puzzle

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up vote
3
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So my last math puzzle was too easy apparently, and this one might be too. Either way, the rules are the same.



  • No rotation.

  • No duplication.

  • No combination.

  • No rounding.

  • No computers.

Good luck to all of you!



You are given the following set of numbers: $1, 2, 3, 5, 6, 9$. Inject these numbers into the following equation to create a true statement.



$$frac21a + frac5bcd + e = f$$



Note



It was pointed out in the chat that I made a typo. I've corrected the number set; and I apologize to everyone for that mistake.






mathematics no-computers







share|improve this question



















  • 2




    Oh boy, 720 possible combinations to choose from... Where to start.....
    – Cubemaster
    5 hours ago






  • 2




    Is there a single solution?
    – Weather Vane
    5 hours ago






  • 2




    You sure there is a solution to this?
    – Robert S.
    4 hours ago






  • 2




    After trying to solve this, I gave up and made a Python script to try every possible combination. I haven't been able to find an answer that doesn't require rounding. I'll check again in case I made a mistake, but it's either that or this requires some lateral thinking.
    – Racso
    4 hours ago






  • 2




    Update from the chat: the 4 in the set should be a 6 instead.
    – Racso
    4 hours ago














up vote
3
down vote

favorite












So my last math puzzle was too easy apparently, and this one might be too. Either way, the rules are the same.



  • No rotation.

  • No duplication.

  • No combination.

  • No rounding.

  • No computers.

Good luck to all of you!



You are given the following set of numbers: $1, 2, 3, 5, 6, 9$. Inject these numbers into the following equation to create a true statement.



$$frac21a + frac5bcd + e = f$$



Note



It was pointed out in the chat that I made a typo. I've corrected the number set; and I apologize to everyone for that mistake.






mathematics no-computers







share|improve this question



















  • 2




    Oh boy, 720 possible combinations to choose from... Where to start.....
    – Cubemaster
    5 hours ago






  • 2




    Is there a single solution?
    – Weather Vane
    5 hours ago






  • 2




    You sure there is a solution to this?
    – Robert S.
    4 hours ago






  • 2




    After trying to solve this, I gave up and made a Python script to try every possible combination. I haven't been able to find an answer that doesn't require rounding. I'll check again in case I made a mistake, but it's either that or this requires some lateral thinking.
    – Racso
    4 hours ago






  • 2




    Update from the chat: the 4 in the set should be a 6 instead.
    – Racso
    4 hours ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











So my last math puzzle was too easy apparently, and this one might be too. Either way, the rules are the same.



  • No rotation.

  • No duplication.

  • No combination.

  • No rounding.

  • No computers.

Good luck to all of you!



You are given the following set of numbers: $1, 2, 3, 5, 6, 9$. Inject these numbers into the following equation to create a true statement.



$$frac21a + frac5bcd + e = f$$



Note



It was pointed out in the chat that I made a typo. I've corrected the number set; and I apologize to everyone for that mistake.






mathematics no-computers







share|improve this question















So my last math puzzle was too easy apparently, and this one might be too. Either way, the rules are the same.



  • No rotation.

  • No duplication.

  • No combination.

  • No rounding.

  • No computers.

Good luck to all of you!



You are given the following set of numbers: $1, 2, 3, 5, 6, 9$. Inject these numbers into the following equation to create a true statement.



$$frac21a + frac5bcd + e = f$$



Note



It was pointed out in the chat that I made a typo. I've corrected the number set; and I apologize to everyone for that mistake.






mathematics no-computers




mathematics no-computers






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago

























asked 5 hours ago









PerpetualJ

2,394229




2,394229







  • 2




    Oh boy, 720 possible combinations to choose from... Where to start.....
    – Cubemaster
    5 hours ago






  • 2




    Is there a single solution?
    – Weather Vane
    5 hours ago






  • 2




    You sure there is a solution to this?
    – Robert S.
    4 hours ago






  • 2




    After trying to solve this, I gave up and made a Python script to try every possible combination. I haven't been able to find an answer that doesn't require rounding. I'll check again in case I made a mistake, but it's either that or this requires some lateral thinking.
    – Racso
    4 hours ago






  • 2




    Update from the chat: the 4 in the set should be a 6 instead.
    – Racso
    4 hours ago












  • 2




    Oh boy, 720 possible combinations to choose from... Where to start.....
    – Cubemaster
    5 hours ago






  • 2




    Is there a single solution?
    – Weather Vane
    5 hours ago






  • 2




    You sure there is a solution to this?
    – Robert S.
    4 hours ago






  • 2




    After trying to solve this, I gave up and made a Python script to try every possible combination. I haven't been able to find an answer that doesn't require rounding. I'll check again in case I made a mistake, but it's either that or this requires some lateral thinking.
    – Racso
    4 hours ago






  • 2




    Update from the chat: the 4 in the set should be a 6 instead.
    – Racso
    4 hours ago







2




2




Oh boy, 720 possible combinations to choose from... Where to start.....
– Cubemaster
5 hours ago




Oh boy, 720 possible combinations to choose from... Where to start.....
– Cubemaster
5 hours ago




2




2




Is there a single solution?
– Weather Vane
5 hours ago




Is there a single solution?
– Weather Vane
5 hours ago




2




2




You sure there is a solution to this?
– Robert S.
4 hours ago




You sure there is a solution to this?
– Robert S.
4 hours ago




2




2




After trying to solve this, I gave up and made a Python script to try every possible combination. I haven't been able to find an answer that doesn't require rounding. I'll check again in case I made a mistake, but it's either that or this requires some lateral thinking.
– Racso
4 hours ago




After trying to solve this, I gave up and made a Python script to try every possible combination. I haven't been able to find an answer that doesn't require rounding. I'll check again in case I made a mistake, but it's either that or this requires some lateral thinking.
– Racso
4 hours ago




2




2




Update from the chat: the 4 in the set should be a 6 instead.
– Racso
4 hours ago




Update from the chat: the 4 in the set should be a 6 instead.
– Racso
4 hours ago










3 Answers
3






active

oldest

votes

















up vote
3
down vote



accepted










$$frac21a + frac5bcd + e = f$$ is true when




a = 2


b = 3


c = 5


d = 9


e = 1


f = 6







share|improve this answer






















  • HAHA That was an awesome loop hole; however, it isn't the correct answer. I've updated the post as I made a typo. That should help lol
    – PerpetualJ
    4 hours ago











  • @Perp Changed now that the 4 is a 6
    – PotatoLatte
    4 hours ago

















up vote
2
down vote













Also not really an answer, but this is the closest I've gotten




$frac21*1 + frac5*439 + 2 = 5tfrac227$ which is only off by $frac227$. So far that's the smallest margin of error that I could find. However rounding isn't allowed so...




EDIT: Actual answer:




$frac21*2 + frac5*359 + 1 = 6$







share|improve this answer






















  • The answer is similar to this. :)
    – PerpetualJ
    4 hours ago










  • @Perp Is mine correct?
    – PotatoLatte
    4 hours ago










  • Dang it! Too slow again! :P
    – Luke C. J. Currie
    4 hours ago






  • 1




    d=9 is an overlap. also f=6 is kinda close to f = 5 and 2/27...
    – Luke C. J. Currie
    3 hours ago










  • He found $d$ and $f$ was close but no rounding allowed.
    – PerpetualJ
    3 hours ago

















up vote
1
down vote













Not an answer, but some facts I have found to help the community:




C != 4,9







share|improve this answer




















  • That is a true statement. :)
    – PerpetualJ
    4 hours ago










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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










$$frac21a + frac5bcd + e = f$$ is true when




a = 2


b = 3


c = 5


d = 9


e = 1


f = 6







share|improve this answer






















  • HAHA That was an awesome loop hole; however, it isn't the correct answer. I've updated the post as I made a typo. That should help lol
    – PerpetualJ
    4 hours ago











  • @Perp Changed now that the 4 is a 6
    – PotatoLatte
    4 hours ago














up vote
3
down vote



accepted










$$frac21a + frac5bcd + e = f$$ is true when




a = 2


b = 3


c = 5


d = 9


e = 1


f = 6







share|improve this answer






















  • HAHA That was an awesome loop hole; however, it isn't the correct answer. I've updated the post as I made a typo. That should help lol
    – PerpetualJ
    4 hours ago











  • @Perp Changed now that the 4 is a 6
    – PotatoLatte
    4 hours ago












up vote
3
down vote



accepted







up vote
3
down vote



accepted






$$frac21a + frac5bcd + e = f$$ is true when




a = 2


b = 3


c = 5


d = 9


e = 1


f = 6







share|improve this answer














$$frac21a + frac5bcd + e = f$$ is true when




a = 2


b = 3


c = 5


d = 9


e = 1


f = 6








share|improve this answer














share|improve this answer



share|improve this answer








edited 4 hours ago

























answered 4 hours ago









PotatoLatte

1,145228




1,145228











  • HAHA That was an awesome loop hole; however, it isn't the correct answer. I've updated the post as I made a typo. That should help lol
    – PerpetualJ
    4 hours ago











  • @Perp Changed now that the 4 is a 6
    – PotatoLatte
    4 hours ago
















  • HAHA That was an awesome loop hole; however, it isn't the correct answer. I've updated the post as I made a typo. That should help lol
    – PerpetualJ
    4 hours ago











  • @Perp Changed now that the 4 is a 6
    – PotatoLatte
    4 hours ago















HAHA That was an awesome loop hole; however, it isn't the correct answer. I've updated the post as I made a typo. That should help lol
– PerpetualJ
4 hours ago





HAHA That was an awesome loop hole; however, it isn't the correct answer. I've updated the post as I made a typo. That should help lol
– PerpetualJ
4 hours ago













@Perp Changed now that the 4 is a 6
– PotatoLatte
4 hours ago




@Perp Changed now that the 4 is a 6
– PotatoLatte
4 hours ago










up vote
2
down vote













Also not really an answer, but this is the closest I've gotten




$frac21*1 + frac5*439 + 2 = 5tfrac227$ which is only off by $frac227$. So far that's the smallest margin of error that I could find. However rounding isn't allowed so...




EDIT: Actual answer:




$frac21*2 + frac5*359 + 1 = 6$







share|improve this answer






















  • The answer is similar to this. :)
    – PerpetualJ
    4 hours ago










  • @Perp Is mine correct?
    – PotatoLatte
    4 hours ago










  • Dang it! Too slow again! :P
    – Luke C. J. Currie
    4 hours ago






  • 1




    d=9 is an overlap. also f=6 is kinda close to f = 5 and 2/27...
    – Luke C. J. Currie
    3 hours ago










  • He found $d$ and $f$ was close but no rounding allowed.
    – PerpetualJ
    3 hours ago














up vote
2
down vote













Also not really an answer, but this is the closest I've gotten




$frac21*1 + frac5*439 + 2 = 5tfrac227$ which is only off by $frac227$. So far that's the smallest margin of error that I could find. However rounding isn't allowed so...




EDIT: Actual answer:




$frac21*2 + frac5*359 + 1 = 6$







share|improve this answer






















  • The answer is similar to this. :)
    – PerpetualJ
    4 hours ago










  • @Perp Is mine correct?
    – PotatoLatte
    4 hours ago










  • Dang it! Too slow again! :P
    – Luke C. J. Currie
    4 hours ago






  • 1




    d=9 is an overlap. also f=6 is kinda close to f = 5 and 2/27...
    – Luke C. J. Currie
    3 hours ago










  • He found $d$ and $f$ was close but no rounding allowed.
    – PerpetualJ
    3 hours ago












up vote
2
down vote










up vote
2
down vote









Also not really an answer, but this is the closest I've gotten




$frac21*1 + frac5*439 + 2 = 5tfrac227$ which is only off by $frac227$. So far that's the smallest margin of error that I could find. However rounding isn't allowed so...




EDIT: Actual answer:




$frac21*2 + frac5*359 + 1 = 6$







share|improve this answer














Also not really an answer, but this is the closest I've gotten




$frac21*1 + frac5*439 + 2 = 5tfrac227$ which is only off by $frac227$. So far that's the smallest margin of error that I could find. However rounding isn't allowed so...




EDIT: Actual answer:




$frac21*2 + frac5*359 + 1 = 6$








share|improve this answer














share|improve this answer



share|improve this answer








edited 4 hours ago

























answered 4 hours ago









Luke C. J. Currie

1,05215




1,05215











  • The answer is similar to this. :)
    – PerpetualJ
    4 hours ago










  • @Perp Is mine correct?
    – PotatoLatte
    4 hours ago










  • Dang it! Too slow again! :P
    – Luke C. J. Currie
    4 hours ago






  • 1




    d=9 is an overlap. also f=6 is kinda close to f = 5 and 2/27...
    – Luke C. J. Currie
    3 hours ago










  • He found $d$ and $f$ was close but no rounding allowed.
    – PerpetualJ
    3 hours ago
















  • The answer is similar to this. :)
    – PerpetualJ
    4 hours ago










  • @Perp Is mine correct?
    – PotatoLatte
    4 hours ago










  • Dang it! Too slow again! :P
    – Luke C. J. Currie
    4 hours ago






  • 1




    d=9 is an overlap. also f=6 is kinda close to f = 5 and 2/27...
    – Luke C. J. Currie
    3 hours ago










  • He found $d$ and $f$ was close but no rounding allowed.
    – PerpetualJ
    3 hours ago















The answer is similar to this. :)
– PerpetualJ
4 hours ago




The answer is similar to this. :)
– PerpetualJ
4 hours ago












@Perp Is mine correct?
– PotatoLatte
4 hours ago




@Perp Is mine correct?
– PotatoLatte
4 hours ago












Dang it! Too slow again! :P
– Luke C. J. Currie
4 hours ago




Dang it! Too slow again! :P
– Luke C. J. Currie
4 hours ago




1




1




d=9 is an overlap. also f=6 is kinda close to f = 5 and 2/27...
– Luke C. J. Currie
3 hours ago




d=9 is an overlap. also f=6 is kinda close to f = 5 and 2/27...
– Luke C. J. Currie
3 hours ago












He found $d$ and $f$ was close but no rounding allowed.
– PerpetualJ
3 hours ago




He found $d$ and $f$ was close but no rounding allowed.
– PerpetualJ
3 hours ago










up vote
1
down vote













Not an answer, but some facts I have found to help the community:




C != 4,9







share|improve this answer




















  • That is a true statement. :)
    – PerpetualJ
    4 hours ago














up vote
1
down vote













Not an answer, but some facts I have found to help the community:




C != 4,9







share|improve this answer




















  • That is a true statement. :)
    – PerpetualJ
    4 hours ago












up vote
1
down vote










up vote
1
down vote









Not an answer, but some facts I have found to help the community:




C != 4,9







share|improve this answer












Not an answer, but some facts I have found to help the community:




C != 4,9








share|improve this answer












share|improve this answer



share|improve this answer










answered 4 hours ago









Cubemaster

1,217228




1,217228











  • That is a true statement. :)
    – PerpetualJ
    4 hours ago
















  • That is a true statement. :)
    – PerpetualJ
    4 hours ago















That is a true statement. :)
– PerpetualJ
4 hours ago




That is a true statement. :)
– PerpetualJ
4 hours ago

















 

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