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Question Number 125080 by ZiYangLee last updated on 08/Dec/20

$$\mathrm{How}\mathrm{many}\mathrm{positive}\mathrm{integers}\mathrm{less}\mathrm{than}$$$$200\mathrm{have}\mathrm{exaxtly}6\mathrm{positive}\mathrm{factors}?$$

Answered by mr W last updated on 08/Dec/20

$$saysuchanumberis$$$$n=p^a{q}^b{r}^c...$$$$withp,q,r,...=prime⩾2$$$$numberoffactors:$$$$(a+1)(b+1)(c+1)...=6=2\times 3$$$$\Rightarrow a=1,b=2,c=d=...=0$$$$\Rightarrow n={pq}^2<200$$$$$$$$\Rightarrow q<\sqrt{\frac{200}{2}}=10$$$$\Rightarrow q=2,3,5,7$$$$q=2:p<\frac{200}{2^2}=50\Rightarrow p=2,3,5,7,...,47\Rightarrow 15numbers$$$$q=3:p<\frac{200}{3^2}=22\Rightarrow p=2,3,5,7,...,19\Rightarrow 8numbers$$$$q=5:p<\frac{200}{5^2}=8\Rightarrow p=2,3,5,7\Rightarrow 4numbers$$$$q=7:p<\frac{200}{7^2}=4\Rightarrow p=2,3\Rightarrow 2numbers$$$$$$$$\Rightarrow totally15+8+4+2=29numbers$$

Commented by mr W last updated on 08/Dec/20

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