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My Wonky CalcuLESS (Posted on 2009-02-18) Difficulty: 3 of 5
I bought a new calculator at the Dollar Store and, sure enough, it’s defective. After some trial and error with it, I discovered that each digit in the display contained the same two pairs of elements (out of the seven elements labeled A to G below) that were somehow ‘cross-wired’. That is, if one element was called upon to illuminate, its partner would illuminate instead. If both were supposed to illuminate, neither would! For example, if A/D and B/F were the faulty pairs, the number 3 would simply display as F/G/C, as illustrated below.

A
F B
G
E C
D


Based on the illuminated elements for each digit given below, find the faulty pairs to then solve the following 3-digit by 2-digit multiplication:

C/DF/B/G/E/CG/E/C
XF/G/C/DF/G/E/C/D
F/E/C/DF/G/C/DF/G/E/C/DF/E/C/D


See The Solution Submitted by rod hines    
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Solution re(2): problems with computer program | Comment 3 of 14 |
(In reply to re: problems with computer program by Robby Goetschalckx)

Actually, the bad elements *are* the same for all digits.
I get the following solution:
 153
x 49
7497

How did I find this ? A lot of information can be gathered by looking at the digits where 5 elements are lit. These must have five elements on in the actual digits, as having all seven elements on would give only three elements lit in the display.
Looking at the middle digit of the top row, this could be a 2,3,5,6 or 9. Looking at each option, trying to find bad elements, there is actually only the 5 left as an option, looking at the corresponding digits below it.
Looking which bad elements belong together, we find the 5,4 and 9 at the position of the tens.
Working in the same manner for the units digit, we also find a solution - the same as for the tens.
Checking this for the other numbers leads to the full solution.

  Posted by Robby Goetschalckx on 2009-02-18 17:48:25

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