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Question Number 124513 by mathocean1 last updated on 03/Dec/20

$$anintegernhas15divisors.nis$$$$divisibleby6butnotdivisibleby8.$$$$determinaten.$$

Commented by mr W last updated on 03/Dec/20

$$324?$$

Commented by soumyasaha last updated on 04/Dec/20

$$\mathrm{N}=2^{\mathrm{p}}.3^{\mathrm{q}}{5}^{\mathrm{r}}..$$$$\mathrm{Considering}1\mathrm{as}\mathrm{not}\mathrm{a}\mathrm{factor}$$$$\mathrm{number}\mathrm{of}\mathrm{factors}=(\mathrm{p}+1)(\mathrm{q}+1)...-1$$$$\mathrm{Now}15=16-1$$$$=4.4-1\mathrm{or}2\times 8-1\mathrm{or}2\times 2\times 4-1$$$$\mathrm{p}+1\ne 4\because \mathrm{no}.\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}8$$$$\therefore \mathrm{p}+1=2\Rightarrow \mathrm{p}=1$$$$\therefore \mathrm{q}=7\mathrm{or}\mathrm{q}=1\mathrm{and}\mathrm{r}=3\mathrm{or}\mathrm{q}=3\mathrm{and}\mathrm{r}=1$$$$$$$$\therefore \mathrm{N}=2^1.3^7\mathrm{or}\mathrm{N}=2^1{3}^1.5^3\mathrm{or}\mathrm{N}=2^1.3^3.5^1$$$$\mathrm{N}=4374\mathrm{or}\mathrm{N}=750\mathrm{or}\mathrm{N}=270$$$$\mathrm{Considering}1\mathrm{as}\mathrm{a}\mathrm{factor}$$$$\mathrm{number}\mathrm{of}\mathrm{factors}=(\mathrm{p}+1)(\mathrm{q}+1)...$$$$\therefore 15=3\times 5\mathrm{or}5\times 3$$$$\mathrm{p}+1\ne 5\because \mathrm{no}.\mathrm{is}\mathrm{not}\mathrm{divisible}\mathrm{by}8$$$$\Rightarrow \mathrm{p}=2\mathrm{and}\mathrm{q}=4$$$$\therefore \mathrm{N}=2^2.3^4=324$$$$$$

Answered by mr W last updated on 04/Dec/20

$$n=p^a{q}^b{r}^c...$$$$(a+1)(b+1)(c+1)...=15=3\times 5$$$$\Rightarrow a=2,b=4$$$$\Rightarrow n=p^2{q}^4$$$$nisdivisibleby6$$$$\Rightarrow nmustcontain{2}^{⩾1}{3}^{⩾1}...\left(1\right)$$$$nisnotdivisibleby8$$$$\Rightarrow nmaycontain{2}^{⩽2}...\left(2\right)$$$$from\left(1\right)and\left(2\right)wecansee$$$$p=2,q=3$$$$\Rightarrow n=2^2{3}^4=324$$

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