X, Y, Z are the sets of all positive divisors | vedclass

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(C) The prime factorizations are:$X = 10^{60} = 2^{60} 	imes 5^{60} \implies n(X) = (60+1)(60+1) = 61^2 = 3721$$Y = 20^{50} = (2^2 	imes 5)^{50} = 2

Vedclass · Accepted Answer

$73001$

Vedclass · Answer

(C) The prime factorizations are:$X = 10^{60} = 2^{60} 	imes 5^{60} \implies n(X) = (60+1)(60+1) = 61^2 = 3721$$Y = 20^{50} = (2^2 	imes 5)^{50} = 2^{100} 	imes 5^{50} \implies n(Y) = (100+1)(50+1) = 101 	imes 51 = 5151$$Z = 30^{40} = (2 	imes 3 	imes 5)^{40} = 2^{40} 	imes 3^{40} 	imes 5^{40} \implies n(Z) = (40+1)^3 = 41^3 = 68921$Intersections:$X \cap Y$: Divisors of $\gcd(10^{60}, 20^{50}) = \gcd(2^{60} 	imes 5^{60}, 2^{100} 	imes 5^{50}) = 2^{60} 	imes 5^{50} \implies n(X \cap Y) = 61 	imes 51 = 3111$$Y \cap Z$: Divisors of $\gcd(2^{100} 	imes 5^{50}, 2^{40} 	imes 3^{40} 	imes 5^{40}) = 2^{40} 	imes 5^{40} \implies n(Y \cap Z) = 41 	imes 41 = 41^2 = 1681$$Z \cap X$: Divisors of $\gcd(2^{40} 	imes 3^{40} 	imes 5^{40}, 2^{60} 	imes 5^{60}) = 2^{40} 	imes 5^{40} \implies n(Z \cap X) = 41^2 = 1681$$X \cap Y \cap Z$: Divisors of $\gcd(2^{60} 	imes 5^{60}, 2^{100} 	imes 5^{50}, 2^{40} 	imes 3^{40} 	imes 5^{40}) = 2^{40} 	imes 5^{40} \implies n(X \cap Y \cap Z) = 41^2 = 1681$Using the Principle of Inclusion-Exclusion:$n(X \cup Y \cup Z) = n(X) + n(Y) + n(Z) - [n(X \cap Y) + n(Y \cap Z) + n(Z \cap X)] + n(X \cap Y \cap Z)$$= 3721 + 5151 + 68921 - [3111 + 1681 + 1681] + 1681$$= 77793 - 6473 + 1681 = 73001$

$X, Y, Z$ are the sets of all positive divisors of $10^{60}, 20^{50},$ and $30^{40}$ respectively. Find $n(X \cup Y \cup Z)$.

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