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 |
I get 9 different cards.
2, 11, 41, 73, 83
9, 15, 21, 33, 51
25, 35, 55, 65, 85
2, 11, 41, 73, 83
9, 21, 33, 51, 69
15, 25, 35, 55, 85
2, 11, 43, 71, 83
9, 15, 21, 33, 51
25, 35, 55, 65, 85
2, 11, 43, 71, 83
9, 21, 33, 51, 69
15, 25, 35, 55, 85
2, 13, 41, 71, 83
9, 15, 21, 33, 51
25, 35, 55, 65, 85
2, 13, 41, 71, 83
9, 21, 33, 51, 69
15, 25, 35, 55, 85
2, 31, 43, 61, 73
9, 15, 21, 33, 51
25, 35, 55, 65, 85
2, 31, 43, 61, 73
9, 21, 33, 51, 69
15, 25, 35, 55, 85
2, 47, 67, 71, 83
9, 21, 33, 51, 69
15, 25, 35, 55, 85
But most of those fail the requirement: “Each column contains one, two or three numbers (at least one of each)”. That is, there must be at least one column with only one number, at least one with two, and at least one with three.
I hadn’t spotted the “at least one of each” stipulation – thanks.
Actually in responding to this comment I realized I’d overlooked the possibility of 9, 21, 33, 51, 69 for the second row. But that again leaves the 40s and 70s columns blank in the last two rows so 2, 31, 43, 61, 73 is the only possible first row. (arthurvause’s second to last solution, which IS valid.) The problem is to find the first row, not the whole card, so there’s only one solution to the stated problem.