## Enigma 1740

Spoiler for New Scientist Enigma 1740: “Sudoprime” (Follow the link to see the puzzle.)

Label the squares:

```A - - - B
C D - E F
- G H I -
J K - L M
N - - - O```

A is 5 and H is 9. C, D, F, I, K, L, M, N, and O must be odd and not 5.

AC must be 53 or 59. Try 59 first:

Then CD must be 97. K cannot be 7 or 9 so must be 1 or 3. Then DGK is 743, 751, or 761. L cannot be 7 or 9 so must be 1 or 3. If DGK is 743 then GHI is 491 and L cannot be 1 or 3. Dead end. If DGK is 751 then GHI is 593 and again L cannot be 1 or 3. Dead end. So DGK is 761, GHI is 691, and L is 3. EIL must be 613. This means O must be 1. F cannot be 1 or 7 or 9 so must be 3; then EF can only be 63 which is not prime. Dead end.

So AC must be 53. CD can be 31 or 37. Try 37:

Same argument as above except we conclude F must be 9, so EF is 69 which is not prime. Dead end.

Then CD is 31. N cannot be 3 or 9 so must be 1 or 7. K, L, and O must be 3 or 7. So LM and MO are 31 and 17 or 71 and 13. Either way M is 1. Try 31 and 17:

Then F is 9. I is 1 or 7. The 3-digit primes ending in 13 or 73 are 113, 173, 313, 373, 613, 673, 773. None is legal for EIL.

So LM and MO are 71 and 13. Then I is 1 or 3. The 3-digit primes ending in 17 or 37 are 137, 317, 337, 617, 937. Only 617 is legal. Then EF is 67. K is 3. GHI is 491 or 691. But 143 is not prime so DGK is 163. B is 4. The only possibilities for JN are 41 and 47, but we already have BF = 47 so JN is 41.

The grid is:

```5 - - - 4
3 1 - 6 7
- 6 9 1 -
4 3 - 7 1
1 - - - 3```

and the answers required are 617 and 41.

## Enigma 1730

Spoiler for New Scientist Enigma 1730: “Prime position” (Follow the link to see the puzzle.)

This one goes quickly, once you realize the 1-digit horizontal and vertical numbers do not count among the 11 answers. Label the blank squares like this:

```2---A
BC-DE
-FGH-
IJ-KL
M---N```

The required primes are BC, DE, FGH, IJ, KL, 2B, AE, CFJ, DHK, IM, and LN (but not 2, A, M, N, or G). All columns have the same sum, namely G, and since CFJ and DHK are prime their digits must not sum to a multiple of 3; therefore G≤8. Prime 2B must be 23 or 29 but 29 gives too large a column sum, therefore B=3. Now I+M≤3 so IM=11 and G=7.

BC=3C=31 or 37, but the latter ruins the column sum. IJ=1J=11 or 13 (not 17). If IJ=13 then CFJ=133 which is not prime, so IJ=11 and CFJ=151, which is prime.

FGH=57H and H<7 so H=1 to give the prime 571. Then D+K=6 and K is odd and not 5, so DHK=313 or 511; only 313 is prime. E and L can only be 1. Now A+N=5 and N is odd. A=2, N=3 doesn’t work (AE=21 isn’t prime) so A=4, N=1. The completed grid is

```2---4
31-31
-571-
11-31
1---1```

and the shaded numbers are 41 and 571.

## Enigma 1724

Spoiler for New Scientist Enigma 1724: “Bingo!” (Follow the link to see the puzzle.)

Start with the bottom row; prime multiples of 5 are 10, 15, 25, 35, 55, 65, 85. If we use 10 then that plus four of the others makes an even sum, which isn’t 5 times a prime, so we must use five of 15, 25, 35, 55, 65, 85. In particular to get 5 times a prime we must use 15, 25, 35, 55, 85 or 25, 35, 55, 65, 85. The ones we must use are 25, 35, 55, and 85.

In the middle row we can use 6, 9, 15, 21, 33, (not 39 because it’s larger than 35), 51, (not 57 because it’s larger than 55), 69, (not 87 because it’s larger than 85). Again we can’t use the even number, 6, because then the sum would be even. So we need five of 9, 15, 21, 33, 51, 69; that is, five of 3 times (3, 5, 7, 11, 17, 23), where the five primes sum to a prime. There are two possibilities: 3(3+5+7+11+17) = 9+15+21+33+51 = 3(43) or 3(3+5+11+17+23) = 9+15+33+51+69 = 3(59). Either way we have to use 15, which means we can’t use 15 in the third row, so the third row must be 25, 35, 55, 65, 85. But that means we can’t use 69 in the second row, so the second row is 9, 15, 21, 33, 51.

In the top row we want the sum to be even, so one of the numbers must be even; we must use 2. We have to have a column with three numbers in it, so we must use 31. Besides these we use three of: 11 or 13; 43 or 47; 61; 71, 73, or 79; 83. There are only three possibilities that add up to a multiple of 30 (that is, a multiple of 2 and 3 and 5), but two of these leave a column blank (we must use a number in the 40s and a number in the 70s). The one remaining possibility is 2, 31, 43, 61, 73.

Here’s the bingo card:

 2 31 43 61 73 9 15 21 33 51 25 35 55 65 85