## St Hubert’s College by Oyler

### Puzzle solution process

This is one possible solving path. The preamble says the college was founded in the 17th century, so 3dn, the year of foundation, must be 16__ (not 1700, if that’s considered part of the 16th century, because that would make 9ac start with zero).

For 5dn, the area of the Great Quad is ≥ 1000 square yards, so the side length is ≥ 32 yards. The perimeter in feet at 6ac is 4×3×(length in yards), so it’s ≥ 4×3×32 = 384. In the grid we have _6_, so the last digit is one of {0, 4, 8} (all multiples of 100 are multiples of 4, so we need the last two digits to be a multiple of 4). The possible multiples of 3 (where the sum of digits is a multiple of 3) are {468, 564, 660, 768, 864, 960}, corresponding to side lengths of {39, 47, 55, 64, 72, 80} yards.

The factors of a number n can always be paired up as (f, n/f), whose product is the number itself, except that the root of a square number, f = n/f, is only counted once; so the number of distinct factors is odd for square numbers and even for non-square numbers. The preamble says the Great Quad side length is a multiple of its number of factors, so from the list above we can eliminate 39, 47, 55 (non-square but they can’t be a multiple of an even number) and 64 (square but with no odd factors, being a power of 2). That leaves 72 with 12 factors {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72} and 80 with 10 {1, 2, 4, 5, 8, 10, 16, 20, 40, 80}, so the area at 5dn is either 722 = 5184 or 802 = 6400. Now 5ac starts with 5 or 6, but the Fibonacci numbers are {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …}, so 5ac = 55, which makes 5dn = 5184, the Great Quad side length = 72 yards, and the perimeter 6ac = 864.

7ac is a multiple of 5ac = 55, so it ends in 5 (because 8dn can’t start with 0). 8dn is 2 times a cube; 2×43 = 128 is too big, and the only two-digit values are {2×23 = 16, 2×33 = 54}, but it must start with 5, so 8dn = 54. Then 10ac is a square starting with 4, ie one of {441, 484} (not 400, which would make 11dn start with 0), and 3dn ends in 1 or 4. 3ac is a prime and must be one of {11, 13, 17, 19}, making 9ac one of {33, 39, 51, 57} respectively, so the third digit of 3dn is 3 or 5. Thus 3dn is restricted to {1631, 1634, 1651, 1654}.

The preamble says that two of the two-digit prime factors of the Master’s wife’s bank code are her age when married and her age now, which must be 24 years apart, because her 25th wedding anniversary is next year. The pairs of primes with a difference of 24 are {(13, 37), (17, 41), (19, 43), (23, 47), (29, 53), (37, 61), (43, 67), (47, 71), (59, 83), (73, 97)}, so the wife’s age now is one of {37, 41, 43, 47, 53, 61, 67, 71, 83, 97}. Of the options for 3dn, the only one with a matching prime factor is 1634 = 2×19×43, so 3dn = 1634, 22ac = 43 and 10ac = 484. That narrows 9ac down to {33, 39} and 3ac to {11, 13}.

Now that we know that the Master’s wife is 43 and she was married at the age of 19, we can deduce her security code. 43×19×13 = 10621 is too big for a four-digit number, so the third factor can only be 11 and the code is 43×19×11 = 8987.

We know 14dn is the product of three two-digit primes, and the clues for 1ac, 2ac and 15ac all say they’re factors of 14dn, so they must be exactly the three prime factors of 14dn. The maximum possible for 14dn is 9993 and the smaller two factors must be at least 11 and 17 (not 11 and 13, because one of those is needed for 3ac), so the maximum for any of its prime factors is 9993/(11×17) ≈ 53.4, ie they’re primes in the range [11, 53]. As the three-digit square 15dn is a multiple of the prime 1ac, it must be 1ac2 times a square number (possibly 1), which limits 1ac to {11, 13, 17, 19, 23, 29, 31} and 15dn is a multiple of one of {121, 169, 289, 361, 529, 841, 961}. But 15ac is ≤ 53, so we can eliminate 841 and 961. For the rest, multiplying by 4 gives a value > 600, except for 4×121 = 484, but 484 is already used at 10ac; therefore 15ac = 1ac2 exactly. The clue for 1ac implies that the current year is ≥ 1900, so 15dn ≥ 1900 − 1634 = 266 and it’s one of {289, 361, 529}.

18ac is 3 times a Fibonacci number, so it’s one of {102, 165, 267, 432, 699} but the middle digit of 15dn is one of {2, 6, 8}, so 18ac is restricted to {267, 699} and 15dn can’t be 289. If 1ac is 23, the maximum for 15ac is 9993/(23×11) ≈ 39.5 but 15dn would be 529, so that’s ruled out. That only leaves 15dn = 361 (so the puzzle was set in 1995) with 1ac = 19 and 18ac = 699, and 15ac is now one of {31, 37}. As 14dn ends in 3 and 1ac ends in 9, the product of 15ac and 2ac must end in 7. If 15ac is 31, the minimum for 14dn is 31×19×17 = 10013, too big, so 15ac = 37 and 2ac ends in 1. It can’t be 31, because 37×19×31 = 21793 is too big, so 2ac = 11 and 14dn = 7733 (= 37×19×11).

Now that 11 is taken, 3ac = 13 and 9ac = 39, making 4dn = 349. Every student is either an undergraduate or a postgraduate, so 4dn = 15ac + 17dn, which makes 17dn = 312 (= 349 − 37).

The length of the Lesser Quad’s perimeter is the snail’s speed multiplied by the time it takes for a lap. 2dn is the speed in feet/hour, so the speed in yards/second is 2dn/(3×3600) and the length in yards is that times 12dn, ie perimeter = 12dn×2dn/10800, so 12dn×2dn = 10800×(perimeter) = 24×33×52×2×(21ac + long side). As 12dn is _79_ it can’t be a multiple of 25 (all multiples of which end in one of {00, 25, 50, 75}), so it can have at most one factor of 5 and the other factor of 5 must come from 2dn, for which we have 15_, so 2dn is one of {150, 155}. If 2dn is 155 = 5×31 then 12dn has to have 25×33×5 = 4320 as a factor (with 31 being a factor of (21ac + long side)), but the multiples of 4320 are {4320, 8640, 12960, …}, none of which match _79_. So 2dn = 150 and the maximum for the perimeter = 12dn×150/10800 is 9799×150/10800 ≈ 136.1 yards. For 11dn, the only number in the range [8000, 8999] that is a multiple of both 5ac = 55 and 150 is 11dn = 8250, making 19ac = 10.

From 1dn starting with 1, the area (in square yards) of the Lesser Quad is ≤ 1999, so 21ac, the shorter side (in yards), is < √1999 ≈ 44.7. We know 21ac ends in 2, so it’s one of {12, 22, 32, 42}. One of the two side lengths has to be a factor of 72 (the Great Quad’s side). If 21ac is 12, itself a factor of 72, then the longer side can be any value between 1010/12 ≈ 84.2 and 1998/12 = 166.5; the smallest possibility gives the perimeter as 2×(12 + 85) = 194, but we know the perimeter has to be ≤ 136, so 21ac isn’t 12. If 21ac is 22, the longer side is a factor of 72 between 1010/22 ≈ 45.91 and 1998/22 ≈ 90.8, ie 72 itself, and the perimeter is 2×(22 + 72) = 188, also too big. If 21ac is 32, the longer side is a factor of 72 between 1010/32 ≈ 31.6 and 1998/32 ≈ 62.4, ie 36, and the perimeter is 2×(32 + 36) = 136. If 21ac is 42, the longer side must be a factor of 72 between 1010/42 ≈ 24.05 and 1998/42 ≈ 47.6, which matches only 36, but that isn’t larger than the short side. Thus the only possibility is 21ac = 32 with a long side of 36, a perimeter of 136, giving 12dn = 9792 (= 136×10800/150) and 20ac = 12. The area is then 1dn = 1152 (= 32×36).

For 7ac we have 5_105 and it’s a multiple of 5ac = 55; the only number that fits is 7ac = 50105 (= 55×911). For 12ac we have 94_ and it’s a multiple of 20ac = 12, ie it’s a multiple of both 3 and 4; to be a multiple of 3 the last digit must be one of {2, 5, 8}, and to be a multiple of 4 it must end with one of {40, 44, 48}; so 12ac = 948.

For 16ac we have _35_7 and the sum of digits is 1ac = 19, so it must be one of {13537, 23527, 33517, 43507}, making 13dn one of {81, 82, 83, 84} respectively. 2dn = 150 = 2×3×52 has 12 factors ({1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150}), as does 84 = 22×3×7 ({1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}), but 13dn must have fewer than 2dn, so 13dn isn’t 84 and 16ac isn’t 43507. Dividing by successive primes, it doesn’t take long to find that 23527 has a factor of 7 and 33517 has a factor of 11, so they’re not primes, which leaves 16ac = 13537 and 13dn = 81, completing the grid.