The number of distinct primes dividing 12! + | vedclass

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Question

(A) Given expression: $12! + 13! + 14!$Factor out $12!$: $12!(1 + 13 + 13 	imes 14)$Simplify the expression inside the parentheses: $12!(1 + 13 + 182

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given expression: $12! + 13! + 14!$Factor out $12!$: $12!(1 + 13 + 13 	imes 14)$Simplify the expression inside the parentheses: $12!(1 + 13 + 182) = 12! 	imes 196$Prime factorization of $196$: $196 = 14^2 = (2 	imes 7)^2 = 2^2 	imes 7^2$Prime factorization of $12!$: $12! = 2^{10} 	imes 3^5 	imes 5^2 	imes 7^1 	imes 11^1$Combining these,$12! 	imes 196 = (2^{10} 	imes 3^5 	imes 5^2 	imes 7^1 	imes 11^1) 	imes (2^2 	imes 7^2) = 2^{12} 	imes 3^5 	imes 5^2 	imes 7^3 	imes 11^1$The distinct prime factors are $2, 3, 5, 7, 11$.Therefore,the number of distinct prime factors is $5$.

The number of distinct primes dividing $12! + 13! + 14!$ is

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