Claude on Question 2, Sep 20

Checking 9973

Since √9973 ≈ 99.9, I only need to test prime divisors up to 99.

2: 9973 is odd → no
3: digit sum = 9+9+7+3 = 28, not divisible by 3 → no
5: doesn't end in 0 or 5 → no
7: 7 × 1424 = 9968, remainder 5 → no
11: alternating sum = 9−9+7−3 = 4 → no
13: 13 × 767 = 9971, remainder 2 → no
17: 17 × 586 = 9962, remainder 11 → no
19: 19 × 524 = 9956, remainder 17 → no
23: 23 × 433 = 9959, remainder 14 → no
29: 29 × 343 = 9947, remainder 26 → no
31: 31 × 321 = 9951, remainder 22 → no
37: 37 × 269 = 9953, remainder 20 → no
41: 41 × 243 = 9963, remainder 10 → no
43: 43 × 231 = 9933, remainder 40 → no
47: 47 × 212 = 9964, remainder 9 → no
53: 53 × 188 = 9964, remainder 9 → no
59: 59 × 169 = 9971, remainder 2 → no
61: 61 × 163 = 9943, remainder 30 → no
67: 67 × 148 = 9916, remainder 57 → no
71: 71 × 140 = 9940, remainder 33 → no
73: 73 × 136 = 9928, remainder 45 → no
79: 79 × 126 = 9954, remainder 19 → no
83: 83 × 120 = 9960, remainder 13 → no
89: 89 × 112 = 9968, remainder 5 → no
97: 97 × 102 = 9894, remainder 79 → no

No prime ≤ 99 divides 9973, so it is prime (in fact, it's the largest four-digit prime).

[Yes]