1
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Sherlock made an elaborate code to remember the 10-digit combination to his locker. This is the code he came up with:

A-K-B-J-C-H-D-G-E-F

Each letter stands for a single digit number (0-9) When two numbers are written together without an operation symbol, they represent a tens digit and a ones digit. Two letters that are written with a slash between them represent a fraction.

D+D+D=F H+H=B G+B+E=F+A+C J+A=CC FxD=KG B÷H=K DxH=CK AxH=KE J÷D=K D/J = H/B

What is the 10-digit combination to Sherlock's locker? Show all your work.

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  • 3
    $\begingroup$ Where did you get this problem from? $\endgroup$ – TheSimpliFire Nov 11 '18 at 15:54
  • $\begingroup$ If I understand properly the puzzle C=1. Is it right? $\endgroup$ – Moti Nov 11 '18 at 16:34
5
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We note that

H + H = 2H = B, so H/B = 1/2. Then D/J = 1/2, so J/D = 2. Hence K = 2. We therefore also have a bunch of doubling pairs: 1-2, 2-4, 3-6, 4-8. But K=2, so J and D are either 3-6 or 4-8. Note that J&D are one set, and H&B are the other distinct set. Further, D and F are a tripled pair. Note that F = 3D, so D = 1,2,3. But from above, D=3,4, so D = 3. Then J = 6 and F = 9. Further, HB must be the other double pair available, 4-8. Hence H = 4 and B = 8. Now, F x D = 9 x 3 = 27 = KG, so G = 7. Since J = 6 and J + A = CC, J+A < 16 so CC = 11 and C = 1. Further, A = 5. Then A x H = KE => 5 x 4 = 2E = 20, so E = 0.

The code is A-K-B-J-C-H-D-G-E-F, which translates to

5-2-8-6-1-4-3-7-0-9.

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3
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A more direct solution:

3D = F. Therefore {D,F} is one of {1,3}, {2,6} or {3,9}.
DxF = KG. We can extend {D,F,KG} = {1,3,03}, {2,6,12} or {3,9,27}.
The first 2 possibilities have duplicate digits, F=G or D=G, they are out.
Remains: K=2, D=3, G=7, F=9.
From there: J/D=K => J/3=2 => J=6
J+A=CC => C=1 and 6+A = 11 => A=5
DxH=CK => 3xH=12 => H=4
AxH=KE => 5x4=2E => E=0
and D/J=H/B => 3/6=4/B => B=8

This translates to the code:

A-K-B-J-C-H-D-G-E-F = 5-2-8-6-1-4-3-7-0-9

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  • $\begingroup$ Thank you for your explanation, +1! But please do be reminded to check others' answers before posting to avoid duplicates. Thanks! $\endgroup$ – Omega Krypton Nov 12 '18 at 0:13
  • $\begingroup$ This is a nice solution as well! I had figured that your starting point was probably faster but that was after I was halfway through. Great job! :D $\endgroup$ – El-Guest Nov 12 '18 at 1:53
  • $\begingroup$ The solution is the same, of course, but the way to arrive at the solution is just as important. $\endgroup$ – Florian F Nov 17 '18 at 18:43
-1
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0 1 2 3 4 5 6 7 8 9
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E C K D H A J G B F

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  • 2
    $\begingroup$ Hi, welcome to puzzling stack exchange! Please put your answer in spoiler tags, and provide an explanation as to how you got the answer. For more information and a badge, take the tour! $\endgroup$ – Zimonze Nov 11 '18 at 16:53

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