This is what the puzzle should look like when you start. If it doesn’t, type “Free” or “Go” in the center square.
Since the free space tells us that the first column increases top to bottom, the highest number (8) has to go in the bottom corner and the lowest number (1) has to go in the top corner.
The clue in the bottom left corner tells us every possible number with a 7 in the ones digit [note: wording changed since the screencaps were made] appears on the card. That means the number 7 has to appear, but since we know that column B increases top to bottom, it has to go right above the 8.
The top left corner tells us that one row is made entirely of numbers whose ones digit is 7. That has to be the fourth row, because no other numbers with a 7 as their ones digit can appear in column B. Column I and column G both have two possible numbers ending in 7 (17&27 and 47&57, respectively), but column N and column O each have only one, so their fourth-row boxes must contain 37 and 67.
Now we turn to the clue in column B, fourth row. Where could the Bingo be? The set of called numbers includes at least one possible number from each column, so the Bingo could be a row or a diagonal; it also includes 5 numbers that would fit in column O. However, we know the free space wasn’t used to win, so the diagonals and the third row are already ruled out. The first, fourth and fifth rows are also ruled out because 1, 7 and 8 have not been called. Finally, by the clue in column O, fourth row, column O contains consecutive odd numbers–which wouldn’t include 62, 70, or 72, the numbers that would be required for column O to be the Bingo. So the Bingo was the second row. The only number that fits in Column B, row 2 is 2. Column N is prime numbers (column N, row 4) so column N, row 2 has to be 41, the only prime number between 31-45 that was called. Column O is odd numbers (column O, row 4) so column O, row 2 has to be 63, since 67 is already taken.
Column N, row 2 says the numbers in column N aren’t in numeric order. Since all the numbers in that column are prime, we know the remaining 2 boxes have to be 31 and 43, the only prime numbers not already appearing on the card. If 43 was in the top row and 31 in the bottom, the numbers would all be in order, so 31 has to go in the top row and 43 in the bottom.
Column N, row 5 tells us only 8 boxes were marked off. You can count the ones we have already placed from the list in Column B, row 4–2, 41, 37, 63, 67, and Free–or you can note that the the little icons for the marked-off squares are Xed out boxes, and just count those! Either way, that tells us that the two remaining marks must be in the Bingo, or Chris wouldn’t have won. But we also know that the number 17 has to appear on the card, because all the numbers whose ones digit is 7 appear (column B, row 1). So 17 must be in column I, row 2. Then column B, row 2 tells us that the Bingo has two even numbers. The only possible Bingo numbers for column G, row 2 are 55 and 56, so column G, row 2 must be 56.
We can quickly place 27, since we know the 4th row numbers all end in 7 and we just placed 17, the other number-ending-in-7 in column I.
Column G, row 2 isn’t adjacent to any numbers containing a 5, but we know that 65 has to appear in column O (column O is consecutive odd numbers, according to column O row 4, and we already have 63 and 67). So 65 has to go in column O, row 5, the only empty box in column O that doesn’t adjoin column G, row 2.
Column O, row 5 isn’t adjacent to any numbers containing a 4. We know column G, row 4 has to be either 47 or 57, since it has to have a 7 as the ones digit, and it can’t be 47 since that has a 4 in it. So column G, row 4 is 57. We know 47 still has to show up, though. It can’t be row 5 for the same reason it can’t be row 4. It also can’t be row 1, because row 1 has to have two even numbers (Column N, row 1), but column B, column N, and column O in row 1 must all be odd numbers (already placed, or the clue from column O, row 4), meaning row 1, column I and column G must both be even–which 47 isn’t. So 47 is in column G, row 3.
We know all the numbers that must be in column O: consecutive odd numbers (column O, row 4), containing at least one number in the 70s (column O, row 2) and the numbers 63, 65 and 67. That means the numbers are 63, 65, 67, 69, and 71. But 69 can’t go in row 3, because column G, row 4 says anything in a straight-line path from that box can’t have a 9, and column O, row 3 is in a straight diagonal line from that box. So 69 is in column O, row 1, and 71 in column O, row 3.
Column O, row 3 says all columns must have either three even numbers or no even numbers. So Column 1, row 3 must be even (since column 1 currently contains 2 even and 2 odd numbers). Also, we know that the number in row 3 has to be between 3 and 6, since we know column 1 is in increasing order (free space). So it could be 4 or 6. But column I, row 4 tells us there are only two pairs of adjoining boxes whose numbers are consecutive, and we already know those pairs: they’re column B, rows 1&2, and column B, rows 4&5. That rules out 6 for column B, row 3, which–with column B, row 4–would make another pair of adjoining boxes with consecutive numbers. So column B, row 3 is 4.
Column O, row 1 says that 3 columns contain the lowest possible number they could contain. Those lowest possible numbers are 1 (column B), 16 (column N), 31 (column N), 46 (column G), and 61 (column O). We have 1 and 31 already, and we know 61 doesn’t appear, which leaves either 16 or 46. But column I, row 2 tells us that 17 is the lowest number in column I, so 16 can’t appear either. That means a 46 must appear. But no number containing a 4 can adjoin column O, row 5, so the 46 must be in column G, row 1.
Column G, row 3 says that one (and only one) column contains numbers with three different tens digits. That column could only be column I (16-30) or column G (46-60), since the possible numbers in the other three columns only have two different tens digits. But column G, row 1 says that the number 60 doesn’t appear. That means the number 30 must appear. Column B, row 3 is adjacent to the largest number in column 2, which has to be 30; the only remaining adjoining box in row I is the third row, so column I, row 3 is 30.
Column I, row 3 says we need two numbers with an 8 as their ones digit. We already have the number 8, so there must be one more. We have open spaces in column I and column G, so we could have an 18, 28, 48, or 58. But we know we can’t have any more adjoining pairs of boxes with consecutive numbers (column I, row 4). Column G, row 5 therefore can’t be 58 (consecutive with 57) or 48 (contains a 4–see column O, row 5), and so the number ending in an 8 must be in column I. But it also can’t be 18, because if it was 18, 18 would be marked off (column B, row 4), and we already have our 8 marked off squares (column N, row 5). So the number ending in 8 is 28, and it can’t be next to the box containing 27, so it’s in column I, row 1.
Column I, row 2 says only one multiple of eleven appears on the card. We don’t have any of those yet, which means it must be a 22 in column I or a 55 in column G (the only multiples of 11 in the number ranges for those columns). But both the remaining boxes must be even numbers (column O, row 3, and both column I and column G currently have two even numbers), so the multiple of 11 is 22, and it goes in column I, row 5.
We only have one box left, and we’re going to have to do it by a process of elimination. We know it must be an even number (column O, row 3), so that means: 46, 48, 50, 52, 54, 56, 58, 60. It can’t contain a 4 (column O, row 5): 50, 52, 56, 58, 60. It can’t be a number that’s consecutive with 57 (column I, row 4): 50, 52, 60. It can’t be 60 (column G, row 1): 50, 52. And it can’t be a multiple of 10 (column I, row 5). So it’s 52! Congratulations!
