Elementary Number Theory by Oyler
Puzzle explanation
The clues left only one ambiguity (41ac = 217 or 237), not four as stated erroneously in the preamble.
This is resolved by recognising that the grid has a Su Doku pattern,
or at least that it's a Latin square.
The statement letters for the two unclued central entries can be arranged to make IODINE,
whose atomic number 53 represents Oyler’s age at the time of publication.
Puzzle solution process
First, we can note the following:
 B means all the digits are odd.
 D numbers match 7n + 2, for some value of n.
 E numbers match 4n + 1 (always odd).
 G (triangular) numbers match n(n + 1) / 2.
 K numbers match 4n + 2 (always even).
 R doesn’t appear in any of the clues, so no entry has repeated digits.
10ac is a twodigit prime (A) whose first digit is three times its second digit (L),
so 10ac = 31.
32ac’s two digits are both odd (B) and the first is three times the second (L);
31 is already used at 10ac, so 32ac = 93.
32dn must now end in 7 (H); its middle digit must be odd (B) and to avoid repeated digits in 32dn and 37ac (not R),
32dn must be 937 or 957.
It can’t be 957, which isn’t prime (A) as it has a factor of 3, so 32dn = 937.
To satisfy H without repeated digits, 37ac = 831.
21ac ends in 5 (F), so 22dn starts with 5 and must end in 4 to satisfy H.
To be a multiple of 4 (O) its middle digit must be even, and to be a multiple of 3 (O again)
the digit sum must be a multiple of 3.
It can’t be 504 (failing P), so 22dn = 564.
38dn is a product of two primes (C), one of which is 5 (F).
Being triangular (G), it has to be either 4 × 5 / 2 = 10 or 5 × 6 / 2 = 15;
10 would make 48ac start with zero, so 38dn = 15.
39dn is a twodigit square (I) with both digits even (J), so 39dn = 64.
42ac is now 54_, one greater than a multiple of 4 (E), so its last digit is 1, 4 or 9.
The digit product of 541 is 20, so it doesn’t satisfy P;
545 is a multiple of 5, but F doesn’t apply to 42ac; therefore 42ac = 549.
39ac is 6_ and prime (A), so it’s 61 or 67, making 34dn either _19 or _79.
The former can’t have a threedigit digit product (P), so 39ac = 67.
33ac is (10x + y) where x and y are even digits (J); 10x is always a multiple of 4, so y must be 2 or 6 to satisfy K.
This makes 34dn 279 or 679; 679 = 7 × 97, satisfying C, which doesn’t apply to 34dn, so 34dn = 279.
33ac can’t be 22 (R doesn’t apply), 42 (failing C) or 62 (L doesn’t apply), so 33ac = 82.
13ac is even (K) so its first digit must be odd (J doesn’t apply).
If it ended in 2 or 6 it would be a multiple of 4, failing K, so it ends in 0, 4 or 8.
9dn must be 3_1 (L and H) so the possibilities are 301 = 7 × 43, 341 = 11 × 31
and 381 = 3 × 127; so, to satisfy Q, 9dn = 381.
18ac is now _1 and the first digit must be even (B doesn’t apply); 41 and 61 are primes
and 81 = 3 × 3 × 3 × 3, so, to satisfy C, 18ac = 21 (3 × 7).
To satisfy M, 13ac must be 38, 58 or 98; 38 = 2 × 19 and 58 = 2 × 29 satisfy C, so 13ac = 98.
4ac is a threedigit, odd (E) square (I).
It can’t be 121, 225 or 441 (R doesn’t apply), or 289, 625 or 841 (M doesn’t apply).
Of the remainder, 169 = 24 × 7 + 1, 361 = 51 × 7 + 4, 529 = 75 × 7 + 4
and 729 = 104 × 7 + 1 all fail D, so 4ac = 961 (137 × 7 + 2)
11ac’s first digit is 3 times its last digit (L) and both are even (J), so it is 6_2.
Since the digits are all different (not R), it must be 602 = 2 × 7 × 43,
642 = 2 × 3 × 107 or 682 = 2 × 11 × 31.
It must have a threedigit prime factor (Q), so 11ac = 642.
4dn is now 94_ and L means that 4dn = 943.
3dn is _6, which J and notR restrict to 26 = 7 × 3 + 5, 46 = 7 × 6 + 4
or 86 = 7 × 12 + 2; only the last matches D, so 3dn = 86.
2dn is a prime (A) with odd digits (B) matching _1; it can’t be 11 (not R)
or 31 (already used at 10ac), so 2dn = 71.
1ac is now _78, its digits are all different (not R) and their sum is prime (M), so it’s 278 or 478.
1dn ends in 5 (F), so it’s 235 or 435; 435 has prime factors of 3, 5 and something else,
so it can’t match C, so 1dn = 235 and 1ac = 278.
14ac is prime (A), so 15dn starts with an odd digit which is 3 times its last digit (L).
It can’t be 3_1 (not H), so it’s 9_3.
The middle digit must be even (not B) and less than 4 (not P).
903 = 3 × 7 × 43 fails C, so 15dn = 923, making 14ac = 59.
16dn ends in 5 (F) so it starts with 4 (H) and 16ac = 43.
41ac ends in 7, so its middle digit must be odd (E) and its first digit must be 2 or 9 (H);
since B doesn’t apply, its first digit must be 2.
From C, 41ac is 217 = 7 × 31 or 237 = 3 × 79, not 257 (prime) or 297 = 3 × 3 × 3 × 11.
From H, 28dn starts with 7 and 28ac ends with 2.
Thus 24dn ends in 2; its first digit is even (J).
It can’t be 22 (not R), 62 (not C) or 82 (already used at 33ac), so 24dn = 42.
28ac is a multiple of 4 and 3 (O), so its middle digit is odd and the digit sum is a multiple of 3;
it can’t be 792 (not P), so 28ac = 732.
36ac is a twodigit, square (I) multiple of 5 (F), so 36ac = 25.
28dn is now 7_52, with the second digit at least 3, to satisfy P and not R.
It can’t be 7352 = 8 × 919 (not Q), 7452 (not O), 7552 or 7752 (not R),
7852 = 52 × 151 (not Q), or 7952 = 2 × 2 × 2 × 2 × 7 × 71
which has more than 12 factors (not N).
So 28dn = 7652.
30ac is two odd digits (B), the second of which is 5 (F).
It can’t be 15 = 3 × 5, 35 = 5 × 7, 55 = 5 × 11
or 95 = 5 × 19 (not C), so 30ac = 75.
31dn is _2_ and its last digit is 3 or 7 (not E).
The sum of its first and last digits is a square (H).
It can’t be 927 (not P), 227 (not R) or 623 (or 31ac would have a repeated digit) so 31dn = 123.
31ac is now 16_ and only 31ac = 164 satisfies M and not R.
40ac ends in 3 and must start with 1 or 6 to satisfy H; it can’t be 1,
or 30dn (7_1) wouldn’t be able to satisfy P, so 40ac is 6_3.
To satisfy M, the middle digit must be 2, 4 or 8; to satisfy P, 40ac = 683.
35ac is even (K) so its first digit is odd (not J).
30dn can’t be 716 (failing P), 736 = 2 × 2 × 2 × 2 × 2 × 23
or 756 = 2 × 2 × 3 × 3 × 3 × 7 (failing Q) or 776 (R),
so 30dn = 796 = 2 × 2 × 199.
To satisfy K and M, 35ac can only be 94 or 98; 98 would give a repeated digit for 27dn, so 35ac = 94.
27dn is now _548; to satisfy M, the first digit must be 2 or 6.
If 26ac is _25 then it can’t satisfy P, so 26ac is _65 and 27dn = 6548.
To satisfy M (and not R), 26ac must be 265 = 5 × 53 or 865 = 5 × 173;
only 26ac = 865 satisfies Q.
20dn’s second digit is even (K) and its first digit is odd (not J) so their sum is odd,
and it’s a square (H), ie, 9.
It can’t be 36 (not I), 72 (not O) or 90 (not F).
If it’s 54, 20ac, a threedigit square (I) which is even (because K makes 6dn even),
must be 576; but that satisfies P, so 20ac can’t be 576, so 20dn isn’t 54.
Therefore 20dn = 18.
Then 20ac can’t be 100 or 144 (not R), so 20ac = 196.
8dn is _92_ and must end in 1, 3 or 7 (A and not R).
Thus 21ac ends in two odd digits, so its first digit must be even (not B) and can’t be 4 or 6,
which would satisfy H for 21ac or 21dn respectively.
To satisfy P, its middle digit must be greater than 1 and the first digit must be greater than 2,
allowing only 835 = 7 × 119 + 2 and 875 = 7 × 125; the latter fails D,
so 21ac = 835 and 8dn is now _923.
7ac ends in 3, so its middle digit must be odd (E).
The first (odd) digit of 8dn must be 5 or 7 (P and not R); 7923 is a multiple of 3, failing A,
so 8dn = 5923 and 7ac is _53, starting with an even digit (not B).
It can’t be 253 = 11 × 23 (not C), 653 (not H) or 853 (not P), so 7ac = 453.
12ac’s two digits are distinct (not R) and both odd (B); from E, the second digit is 3 or 7.
It’s not prime (not A), which leaves only 57 and 93.
It can’t be 93 (not L), so 12ac = 57.
Then 7dn is 47_ with the last digit even (M) and greater than 4 (P and not R).
It can’t be 478 = 2 × 239 (N and not Q), so 7dn = 476.
6dn is 15_6 with the third digit even (K), greater than 3 (P) and not 6 (not R), ie 1546 or 1586.
17ac is now _46 or _86 with the first digit odd (not J), greater than 3 (P and not H) and not 9 (not M).
The possibilities are 546 = 78 × 7, 586 = 83 × 7 + 5, 746 = 106 × 7 + 4
and 786 = 112 × 7 + 2, of which only 17ac = 786 satisfies D, making 6dn = 1586.
19dn ends in 8 and its second digit must be odd (K), so its first digit must be even (M).
It can’t be 8 or 2, which would give a repeated digit in either 19dn or 19ac, so they start with 4 or 6.
For M, 19ac can be 421, 425, 427, 623, 625 or 629; since E doesn’t apply,
it can only be 427 or 623; since H doesn’t apply, 19ac = 427.
19dn is now 4_8, and since P doesn’t apply, the (odd) middle digit must be 1 or 3.
438 doesn’t match M, so 19dn = 418.
16dn is now 47_5.
To avoid satisfying P, it must be 4705, 4785 or 4795.
It can’t be 4705 (not E) or 4785 = 3 × 5 × 11 × 29 (failing Q), so 16dn = 4795.
To satisfy K, the middle digit of 29ac is odd.
To satisfy E, the third digit of 21dn is also odd and it’s not 1 (which would make 29ac too small to satisfy Q),
3 (because R doesn’t apply to 21dn) or 5 (because H doesn’t apply to 29ac).
To avoid satisfying P, if the first digit of 29ac is 7 the middle digit must be less than 4,
and if the first digit is 9 the middle digit must be less than 3; of the remaining possibilities,
714 = 2 × 3 × 7 × 17 fails Q, so 29ac is 734 or 914.
25dn can’t start with 1 (not H) or 8 (not R).
If its middle digit is 3, the first digit must be less than 5 to avoid satisfying P;
it can’t be 238 (not M), 338 (not R) or 438 (fails Q).
Therefore its middle digit is 1.
It can’t be 218 or 418 (not M) or 618 (not D).
Of the remainder, only 718 = 2 × 359 satisfies Q, so 25dn = 718 and 29ac = 914.
21dn is 8_93.
To satisfy P, the second digit must be 1, 2 or 4 (not 3, which would give a repeated digit).
Only 21dn = 8293 is a prime, satisfying A.
To complete the Latin square, 41ac is chosen to be 417 instead of 437 and the central square is filled with a 5.
This makes 23ac = 139854276, which satisfies D, I, N and O, and 5dn = 627953481, which satisfies E and I.
These letters make IODINE, the element referred to in the title.
