46
$\begingroup$

I display to you a set of 5 equations that cannot be solved*.

I want you to find the solution:


B + C = 1
B + C + F + G = 4
A + B + C = 7
A + B + C + D + G = 3
A + B + E + D + G = 2


Time’s ticking, don't wait around!

What is the solution to the above equations?


*cannot be solved as in, the equations do not have a unique solution.

$\endgroup$
17
  • 4
    $\begingroup$ If they cannot be solved, why do you want us to find the solutions? :P . $\endgroup$ – Anonymous Dec 30 '20 at 15:27
  • 2
    $\begingroup$ The answer given by user39583 is definitely nicer but as it stands the system of equations can be solved and has more than one solution. Say A=6, B=2, C=-1, D=-6, E=-2, F=1 and G=2 is a solution if I didn't screw up my algebra. $\endgroup$ – quarague Dec 31 '20 at 11:18
  • 2
    $\begingroup$ @quarague very true, I have a feeling even if I wrote out the ‘equations’ for every number that can be made there would still technically be a real solution. I’ll just claim that the fact there are multiple solutions means there’s not an ‘exact’ solution :) $\endgroup$ – Beastly Gerbil Dec 31 '20 at 11:46
  • 2
    $\begingroup$ @smci unique is a much better word there I agree. The question is supposed to be quite vague on purpose, but I will change to ‘these equations don’t have a unique solution'. Saying they aren’t supposed to be solved numerically gives too much away $\endgroup$ – Beastly Gerbil Jan 1 at 13:56
  • 2
    $\begingroup$ @smci I don’t see how the title is misleading. If anything it is more truthful than anything. I have stated that the equations aren’t solvable, said what I mean by solvable, and it is true. There is a solution, but not in the traditional sense. Clickbait is when you title it something completely false or unrelated. This title is both related and true. It is definitely more than ‘reasonably accurate’, and I haven’t seen anyone else have any issues, so will keep it as is. It is also nearly no longer on HNQ anyways if that’s the issue $\endgroup$ – Beastly Gerbil Jan 2 at 2:53
50
$\begingroup$

All the hints point us towards

Seven-segment display. The letters correspond to the segments needed to display each of the numbers.

enter image description here

(Image from the wikipedia page)

$\endgroup$
6
  • $\begingroup$ Correct! Good spot :) And I probably overdid it slightly with the hints, but good spot seeing the hints too! $\endgroup$ – Beastly Gerbil Dec 30 '20 at 15:55
  • $\begingroup$ @BeastlyGerbil - You should accept this answer as correct. $\endgroup$ – Vilx- Jan 1 at 0:14
  • 2
    $\begingroup$ @Vilx- don’t worry I will! I usually leave answers a couple of days before accepting that’s all, but this answer will definitely get the check! $\endgroup$ – Beastly Gerbil Jan 1 at 0:21
  • $\begingroup$ @BeastlyGerbil - I don't see the point (especially since you can always change the checkmark if you feel like it), but OK. Fair enough. :) $\endgroup$ – Vilx- Jan 1 at 0:54
  • 2
    $\begingroup$ @Vilx- I could be wrong, but puzzles seem to perform better when you don’t accept straight away, and think it affects HNQs too. I make sure to always accept for all my puzzles though :) $\endgroup$ – Beastly Gerbil Jan 1 at 0:55
6
$\begingroup$

This question is really weird and has some weird ambiguities: I'm not sure if I did it right, so please critique my logic! I also don't know how to make the whole thing a spoiler so they are kind of haphazardly placed.

We know that

  1. B + C = 1
  2. B + C + F + G = 4
  3. A + B + C = 7
  4. A + B + C + D + G = 3
  5. A + B + E + D + G = 2

From 1 and 3 we know that

A = 6

From 1 and 2 we know that

F + G = 3

so

G = F - 3

From 3 and 4 we know that

D + G = -4

so

D = -4 - G

so substituting the above equation for G we get

D = F - 3 - 4

so

D = F- 7

and

F = 7 + D

We can substitute the above into equation 2 and 4 to get

B + C + (7 + D) + (3 - G) = 4

so

D = -7

and

A + B + C + (-7) + G = 3

so

G = 3

and substituting this back into equation 2 gives us

F = 0

This is where things get weird:

We know from equations 4 and 5 that

C - E = 1

and using that in equation 1 we find that

C = -B + 1

therefore

B = -E

This means that we can use literally any value for B, and as long as C and E follow the above conventions, all the equations will still be true.

$\endgroup$
2
  • 1
    $\begingroup$ There's a mistake where you substitute your expression for G; you should substitute (F - 3) so that way D = -4 - G = -4 - (F - 3) = -4 - F + 3 = -1 - F, not F - 7 $\endgroup$ – bobble Dec 31 '20 at 17:22
  • $\begingroup$ Interesting, the maths is slightly off at the start, but I imagine the result will still be the same. Quite happy in that case as I didn't want there to be an actual 'solution' rather the solution be the other answer, so quite happy its worked out this way :) +1 for trying to find this answer though! $\endgroup$ – Beastly Gerbil Dec 31 '20 at 17:44
1
$\begingroup$

I don't know if there's more than one solution, but my calculations led to:

A = 6
B = 0
C = 1
D = -4
E = 0
F = 3
G = 0

$\endgroup$
1
  • 2
    $\begingroup$ This a solution, but there multiple (possible infinite) amount of exact solutions $\endgroup$ – Beastly Gerbil Jan 1 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.