If n=(210)^2(360)(143),then the total number | vedclass

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(D) The given number is $n = (210)^2(360)(143)$.First,we find the prime factorization of $n$:$n = (2 	imes 3 	imes 5 	imes 7)^2 	imes (2^3 	imes

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$1438$

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(D) The given number is $n = (210)^2(360)(143)$.First,we find the prime factorization of $n$:$n = (2 	imes 3 	imes 5 	imes 7)^2 	imes (2^3 	imes 3^2 	imes 5) 	imes (11 	imes 13)$$n = (2^2 	imes 3^2 	imes 5^2 	imes 7^2) 	imes (2^3 	imes 3^2 	imes 5) 	imes (11 	imes 13)$$n = 2^{2+3} 	imes 3^{2+2} 	imes 5^{2+1} 	imes 7^2 	imes 11^1 	imes 13^1$$n = 2^5 	imes 3^4 	imes 5^3 	imes 7^2 	imes 11^1 	imes 13^1$The total number of factors of $n$ is given by the product of (exponent + $1$) for each prime factor:Total factors $= (5+1)(4+1)(3+1)(2+1)(1+1)(1+1)$Total factors $= 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 2 = 1440$The trivial factors are $1$ and $n$ itself.Therefore,the number of non-trivial factors is $1440 - 2 = 1438$.Hence,option $D$ is correct.

If $n=(210)^2(360)(143)$,then the total number of non-trivial factors of $n$ is

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