From FiveThirtyEight’s “Riddler” feature: fill in this table with single-digit numbers, such that the entries in the margins are the products of the corresponding entries in the interior of the table.

294 | |||

216 | |||

135 | |||

98 | |||

112 | |||

84 | |||

245 | |||

40 | |||

8890560 | 156800 | 55566 |

We start by factorizing the column products, to get , , and respectively. Since the second-column product isn’t divisible by 3, the second column must consist of only 1, 2, 4, 5, 7, and 8. The third column isn’t divisible by 4 or 5 so it can’t contain 4, 5, or 8; furthermore it contains only a single even number (2 or 6).

We can explicitly enumerate the possibilities for each row. For example for the row with product 84 we have

```
expand.grid(1:9, 1:9, 1:9) %>% filter(8890560 %% Var1 == 0 & 156800 %% Var2 ==0 & 55566 %% Var3 == 0) %>%
mutate(prod = Var1 * Var2 * Var3) %>% filter(prod == 84)
```

which returns the data frame

```
Var1 Var2 Var3 prod
1 6 7 2 84
2 7 4 3 84
3 4 7 3 84
4 7 2 6 84
5 2 7 6 84
6 6 2 7 84
7 3 4 7 84
```

and so there are seven possibilities for this row. 7, 6, 2 doesn’t appear because the second column can’t contain a multiple of three.

This is actually enough to fill in a few entries, and we also can list all the possibilities for the remaining ones:

6, 7 | 7 | 6, 7 | 294 |

3, 6, 9 | 4, 8 | 3, 6, 9 | 216 |

3, 9 | 5 | 3, 9 | 135 |

2, 7 | 2, 7 | 2, 7 | 98 |

2, 4, 7, 8 | 2, 4, 7, 8 | 2, 7 | 112 |

2, 3, 4, 6, 7 | 2, 4, 7 | 2, 3, 6, 7 | 84 |

5, 7 | 5, 7 | 7 | 245 |

4, 5, 8 | 4, 5, 8 | 1, 2 | 40 |

8890560 | 156800 | 55566 | |

Now the first column has product , so it must have a single 5 and three 7s. The remaining four entries have to multiply to so they must be two 9s and two 8s. That lets us complete the first column, because there is only two possible locations for a 9, two for an 8, and one for a 5. And knowing those first-column values allows us to complete various rows:

7 | 7 | 6 | 294 |

9 | 4, 8 | 3, 6 | 216 |

9 | 5 | 3 | 135 |

7 | 2, 7 | 2, 7 | 98 |

8 | 2, 7 | 2, 7 | 112 |

7 | 2, 4 | 3, 6 | 84 |

5 | 7 | 7 | 245 |

8 | 5 | 1 | 40 |

8890560 | 156800 | 55566 | |

The third column only contains a single even number, which is enough to finish that column, and then work out the second column by arithmetic:

7 | 7 | 6 | 294 |

9 | 8 | 3 | 216 |

9 | 5 | 3 | 135 |

7 | 2 | 7 | 98 |

8 | 2 | 7 | 112 |

7 | 4 | 3 | 84 |

5 | 7 | 7 | 245 |

8 | 5 | 1 | 40 |

8890560 | 156800 | 55566 | |

I was honestly surprised this puzzle was solvable – I didn’t believe there was enough information at first. I think it works out because the first-column product 8890560 is large enough that we can determine uniquely what the values in the column are and only have to put them in order; the third-column only having one even value works as well.

Also, I believe this puzzle was part of the MIT Mystery Hunt that took place this weekend (which I haven’t competed in in a Very Long Time). The Riddler column was named “Can You Hunt For The Mysterious Numbers?” and it says the puzzle was by Barbara Yew, and googling that “name” finds an MIT web page at yewlabs.mit.edu for something called “MYST2021: Maturing Young Scientific Theories: Expanding Reality & You” – the first letters spell “MYSTERY”.