Question

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

• A. 70301
• B. 30701
• C. 73001
• D. 70031

Answer 1

Answer: (C)

$x=2\times 5^{60};n(x)=6!\times 5!$
$n(y)=10!\times 5!,wherey-2^{100}\times 5^{10}$
$z=2^{40}\times 3^{40}\times 5^{40}$
$n(z)=41^3$
$n(x\cap y)=6!\times 5!$
$n(x\cap z)=41^2=n(z\cap x)$
$n(x\cap y\cap z)=41^2$
$n(x\cup y\cup z)=n(x)+n(y)+n(z)$
$=n(x\cap y)-(y\cap z)-n(z\cap x)+n(x\cap y\cap z)$
$=61^2+101\times 51+(41{)}^3-61\times 51-41^2-41^2$
$=61(61-51)+41^2(41-1)+101\times 51=73001$