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If $F(n)$ denotes the set of all divisors of $n$ except 1, what is the least value of y satisfying $[F(20) \cap F(16)] \subseteq F(y$
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The correct answer is:
4
Given that $\mathrm{F}(\mathrm{n})=$ set of all divisors of $\mathrm{n}$ except $\mathrm{F}(20)=\{2,4,5,10,20\}$
and $\mathrm{F}(16)=\{2,4,8,16\}$
$\mathrm{F}(20) \cap \mathrm{F}(16)=\{2,4,5,10,20\} \cap\{2,4,8,16\}$
$=\{2,4\}$
Also, $\{\mathrm{F}(20) \cap \mathrm{F}(16)\} \subseteq \mathrm{F}(\mathrm{y})$
So, least value of $y=2$
and $\mathrm{F}(16)=\{2,4,8,16\}$
$\mathrm{F}(20) \cap \mathrm{F}(16)=\{2,4,5,10,20\} \cap\{2,4,8,16\}$
$=\{2,4\}$
Also, $\{\mathrm{F}(20) \cap \mathrm{F}(16)\} \subseteq \mathrm{F}(\mathrm{y})$
So, least value of $y=2$
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