Enigma 1742

Spoiler for New Scientist Enigma 1742: “Chip-chop” (Follow the link to see the puzzle.)

This is pretty straightforward. Label the digits ABCDEFG. The following sums must be equal to 0 mod 3 (but not 0 mod 9):

  1. A+B+C+D+E+F
  2. B+C+D+E+F+G

and the following must not be equal to 0 mod 3:

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

The seven digits are consecutive. For all the numbers to be odd, D through G must be odd digits; four of seven digits odd means we can only use 1–7 or 3–9. But in the latter case sum 3 = 3+4+5+6+7+8+9 is 42 so this is ruled out.

So we are using 1–7 and sum 3 = 1+2+3+4+5+6+7 = 28. Then G = 7 and sum 1 = 28–7 = 21 is the only possibility. (28–3 and 28–5 are ≠ 0 mod 3, and 28–1 = 0 mod 9.) For sum 2 the only possibility is A = 4, 28–4 = 24.

Now sum 4 = 28–4–7 = 17. Possible values for F are 1 or 3 but not 5 to make sum 5 = 16 or 13 but not 12. If F is 3 then and E are 1 and 5, but then sum 9 is 6. So F is 1. Now D and E are 3 and 5. D can’t be 3 (sum 10) so D = 5, E = 3.

B and are 2 and 6. For sum 6 to be ≠ 0 mod 3, must be 2. The seven digit number then is 4625317 and the sums are:

  1. A+B+C+D+E+= 21
  2. B+C+D+E+F+G = 24
  3. A+B+C+D+E+F+G = 28
  4. B+C+D+E+= 17
  5. B+C+D+E = 16
  6. C+D+E+= 11
  7. C+D+= 10
  8. C+= 7
  9. D+= 8
  10. = 5
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