Number TheoryChallengeFactors & Multiples ShareFactors & Multiples · Challenge problemWrite 100!100!100! as a product of primes, 100!=2a23a35a5⋯97a97100! = 2^{a_2} 3^{a_3} 5^{a_5} \cdots 97^{a_{97}}100!=2a23a35a5⋯97a97. Find the sum of all the exponents, a2+a3+a5+⋯+a97a_2 + a_3 + a_5 + \cdots + a_{97}a2+a3+a5+⋯+a97.Check Show answer Walk through it step by step Practice more like this Review this concept