AMC 10A/12A 2024 (Problem 5/4)What is the least value of nnn such that n!n!n! is a multiple of 202420242024?(A) 11\text{(A)}\;11(A)11(B) 21\text{(B)}\;21(B)21(C) 22\text{(C)}\;22(C)22(D) 23\text{(D)}\;23(D)23(E) 253\text{(E)}\;253(E)253Check AnswerYour answer:ABCDECheckHints (4)Hint 1Prime factorize 202420242024.Hint 22024=23×11×232024=2^3\times11\times232024=23×11×23Hint 3n!n!n! should be a multiple of 232323.Hint 4By Hint 3, n≥23n\ge23n≥23 and 23!23!23! works sinceit is also multiple of 23=82^3=823=8 and 111111.Final Answer(D) 23