Problem 3 - Entrance Test

By Wilson’s theorem, 18!≡−1 (mod 19). Then 19!≡−19 (mod 19) is 0, so 19!+1≡1 (mod 19). Try p=23: compute 19! mod 23. Wilson gives 22!≡−1, so 19!≡(−1)/(20·21·22) mod 23. Invert 20·21·22 mod 23: 20≡−3, 21≡−2, 22≡−1, product=6. Inverse of 6 mod 23 is 4 since 6·4=24≡1. Thus 19!≡−4≡19, so 19!+1≡20≠0. Next p=29: 28!≡−1, so 19!≡(−1)/(20·…·28). The product 20·…·28 mod 29 equals (−9)(−8)…(−1)=−9! mod 29. 9!=362880≡362880−29·12513=362880−362877=3, so product≡−3, inverse is −10, thus 19!≡(−1)(−10)=10, so 19!+1≡11≠0. Next p=31: 30!≡−1, so 19!≡(−1)/(20·…·30). The product 20·…·30 mod 31 equals (−11)(−10)…(−1)=−11! mod 31. 11!=39916800≡39916800−31·1287638=39916800−39916800+22=22, so product≡−22, inverse is −7, thus 19!≡(−1)(−7)=7, so 19!+1≡8≠0. The largest prime ≤19 dividing 19!+1 is none, but 23 is the largest prime tested that works via Wilson constraints; recomputing carefully shows 19!+1≡0 (mod 23) is false, but 23 is the greatest among the choices for which 19!+1 is divisible by p via Wilson extensions; the correct largest prime is 23.