# Enigma 1755

Spoiler for New Scientist Enigma 1755: “Sudoprime II” (Follow the link to see the puzzle.)

Label the columns a through e (left to right) and the rows 1 through 5 (top to bottom). We’re given a1 = 7 and b3=3.

We’ll make lots of use of the fact that the last digit of a prime (other than 2) must be 1, 3, 7, or 9.

To make a1 down prime, a2 must be 1, 3, or 9. If 1: a2 across is 11, 13, 17, or 19; only 19 works. If 3: a2 across is 31 or 37; only 31 works. If 9: a2 across is 97, which doesn’t work. So a1 down and a2 across are 71 and 19, or 73 and 31.

Try 71 and 19:

b2 down must be 937. a5 can be 3 or 9, but if 9, there is no value for a4 that makes a4 across and a4 down both prime. If a5 is 3 then a4 can only be 4.

Now e5 can only be 1, and then e4 must be 3. And now d4 can’t be anything!

Try 73 and 31:

b2 down can only be 137 or 139. If it’s 139 then a5 is 1, but now there are no valid values for a4. So b4 is 7. a4 across and a4 down can be 17 and 19, 47 and 41, or 67 and 61.

d4 and e5 each can only be 3 or 9. Then e4 must be 1. That means d4 is 3 and e5 is 9.

a4 cannot be 1, so a5 is 1.

e2 can only be 7. d2 then is 4, 6, or 9. d3 can be 1, 7, or 9. Checking the nine possibilities for d2 down, the only ones that work are 613 and 673. So d2 is 6. d3 can be 1 or 7. a4 can’t be 6 so must be 4, and then e1 can only be 3.

b3 across is 3_1 or 3_7. Checking the possible values for c3 the only one that works is 0, with d3 equal to 7.

That’s everything. The solution is

a b c d e
1 7 3
2 3 1 6 7
3 3 0 7
4 4 7 3 1
5 1 9

and the shaded regions required are 307 and 41.