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.

$$\frac{21a + \frac{5b}{c}}{d} + e = f$$


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.

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16
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    $\begingroup$ Oh boy, 720 possible combinations to choose from... Where to start..... $\endgroup$
    – Cubemaster
    Sep 21, 2018 at 19:14
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    $\begingroup$ Is there a single solution? $\endgroup$ Sep 21, 2018 at 19:16
  • 2
    $\begingroup$ There is only one solution. $\endgroup$ Sep 21, 2018 at 19:19
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    $\begingroup$ You sure there is a solution to this? $\endgroup$
    – Robert S.
    Sep 21, 2018 at 19:23
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    $\begingroup$ 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. $\endgroup$
    – Racso
    Sep 21, 2018 at 19:55

3 Answers 3

4
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$$\frac{21a + \frac{5b}{c}}{d} + e = f$$ is true when

a = 2
b = 3
c = 5
d = 9
e = 1
f = 6

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2
  • $\begingroup$ 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 $\endgroup$ Sep 21, 2018 at 20:13
  • $\begingroup$ @Perp Changed now that the 4 is a 6 $\endgroup$ Sep 21, 2018 at 20:14
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Also not really an answer, but this is the closest I've gotten

$\frac{21*1 + \frac{5*4}{3}}{9} + 2 = 5\tfrac{2}{27}$ which is only off by $\frac{2}{27}$. So far that's the smallest margin of error that I could find. However rounding isn't allowed so...

EDIT: Actual answer:

$\frac{21*2 + \frac{5*3}{5}}{9} + 1 = 6$

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5
  • $\begingroup$ The answer is similar to this. :) $\endgroup$ Sep 21, 2018 at 19:48
  • $\begingroup$ @Perp Is mine correct? $\endgroup$ Sep 21, 2018 at 20:05
  • $\begingroup$ Dang it! Too slow again! :P $\endgroup$ Sep 21, 2018 at 20:22
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    $\begingroup$ d=9 is an overlap. also f=6 is kinda close to f = 5 and 2/27... $\endgroup$ Sep 21, 2018 at 20:33
  • $\begingroup$ He found $d$ and $f$ was close but no rounding allowed. $\endgroup$ Sep 21, 2018 at 20:34
1
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Not an answer, but some facts I have found to help the community:

C != 4,9

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1
  • $\begingroup$ That is a true statement. :) $\endgroup$ Sep 21, 2018 at 19:47

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