*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):

*A*+*B*+*C*+*D*+*E*+*F**B*+*C*+*D*+*E*+*F**+G*

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

*A*+*B*+*C*+*D*+*E*+*F**+G**B*+*C*+*D*+*E*+*F**B*+*C*+*D*+*E**C*+*D*+*E*+*F**C*+*D*+*E**C*+*D**D*+*E**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 *D *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 *C *are 2 and 6. For sum 6 to be ≠ 0 mod 3, *C *must be 2. The seven digit number then is 4625317 and the sums are:

*A*+*B*+*C*+*D*+*E*+*F*= 21*B*+*C*+*D*+*E*+*F**+G*= 24*A*+*B*+*C*+*D*+*E*+*F**+G*= 28*B*+*C*+*D*+*E*+*F*= 17*B*+*C*+*D*+*E*= 16*C*+*D*+*E*+*F*= 11*C*+*D*+*E*= 10*C*+*D*= 7*D*+*E*= 8*D*= 5