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

an integer n has 15 divisors. n is divisible by 6 but not divisible by 8. determinate n.

$${an}\:{integer}\:{n}\:{has}\:\mathrm{15}\:{divisors}.\:{n}\:{is} \\ $$$${divisible}\:{by}\:\mathrm{6}\:{but}\:{not}\:{divisible}\:\:{by}\:\mathrm{8}. \\ $$$${determinate}\:{n}. \\ $$

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

324?

$$\mathrm{324}? \\ $$

Commented by soumyasaha last updated on 04/Dec/20

N = 2^p .3^q 5^r .. Considering 1 as not a factor number of factors = (p+1)(q+1)... −1 Now 15 = 16 −1 = 4.4 −1 or 2×8 −1 or 2×2×4−1 p +1 ≠ 4 ∵ no. is not divisible by 8 ∴ p+1 = 2 ⇒ p = 1 ∴ q = 7 or q = 1 and r = 3 or q=3 and r=1 ∴ N = 2^1 .3^7 or N = 2^1 3^1 .5^3 or N = 2^1 .3^3 .5^1 N = 4374 or N = 750 or N = 270 Considering 1 as a factor number of factors = (p+1)(q+1)... ∴ 15 = 3×5 or 5×3 p +1 ≠ 5 ∵ no. is not divisible by 8 ⇒ p=2 and q=4 ∴ N = 2^2 .3^4 = 324

$$\:\mathrm{N}\:=\:\mathrm{2}^{\mathrm{p}} .\mathrm{3}^{\mathrm{q}} \mathrm{5}^{\mathrm{r}} .. \\ $$$$\:\:\mathrm{Considering}\:\mathrm{1}\:\mathrm{as}\:\mathrm{not}\:\mathrm{a}\:\mathrm{factor} \\ $$$$\:\:\mathrm{number}\:\mathrm{of}\:\mathrm{factors}\:=\:\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{q}+\mathrm{1}\right)...\:−\mathrm{1} \\ $$$$\:\:\mathrm{Now}\:\mathrm{15}\:=\:\mathrm{16}\:−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}.\mathrm{4}\:−\mathrm{1}\:\:\:\:\mathrm{or}\:\:\:\mathrm{2}×\mathrm{8}\:−\mathrm{1}\:\:\:\mathrm{or}\:\mathrm{2}×\mathrm{2}×\mathrm{4}−\mathrm{1} \\ $$$$\:\:\:\mathrm{p}\:+\mathrm{1}\:\neq\:\mathrm{4}\:\because\:\mathrm{no}.\:\mathrm{is}\:\mathrm{not}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{8} \\ $$$$\:\:\therefore\:\mathrm{p}+\mathrm{1}\:=\:\mathrm{2}\:\:\Rightarrow\:\mathrm{p}\:=\:\mathrm{1} \\ $$$$\:\:\therefore\:\mathrm{q}\:=\:\mathrm{7}\:\:\:\mathrm{or}\:\:\:\:\mathrm{q}\:=\:\mathrm{1}\:\mathrm{and}\:\mathrm{r}\:=\:\mathrm{3}\:\mathrm{or}\:\:\mathrm{q}=\mathrm{3}\:\mathrm{and}\:\mathrm{r}=\mathrm{1} \\ $$$$\:\:\:\: \\ $$$$\:\:\therefore\:\mathrm{N}\:=\:\mathrm{2}^{\mathrm{1}} .\mathrm{3}^{\mathrm{7}} \:\:\mathrm{or}\:\:\:\mathrm{N}\:=\:\mathrm{2}^{\mathrm{1}} \mathrm{3}^{\mathrm{1}} .\mathrm{5}^{\mathrm{3}} \:\:\mathrm{or}\:\:\:\mathrm{N}\:=\:\mathrm{2}^{\mathrm{1}} .\mathrm{3}^{\mathrm{3}} .\mathrm{5}^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\mathrm{N}\:=\:\mathrm{4374}\:\:\mathrm{or}\:\:\mathrm{N}\:=\:\mathrm{750}\:\:\mathrm{or}\:\:\mathrm{N}\:=\:\mathrm{270} \\ $$$$\:\:\mathrm{Considering}\:\mathrm{1}\:\mathrm{as}\:\mathrm{a}\:\mathrm{factor} \\ $$$$\:\:\mathrm{number}\:\mathrm{of}\:\mathrm{factors}\:=\:\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{q}+\mathrm{1}\right)... \\ $$$$\:\:\:\:\therefore\:\:\mathrm{15}\:=\:\mathrm{3}×\mathrm{5}\:\:\mathrm{or}\:\:\mathrm{5}×\mathrm{3} \\ $$$$\:\:\:\mathrm{p}\:+\mathrm{1}\:\neq\:\mathrm{5}\:\because\:\mathrm{no}.\:\mathrm{is}\:\mathrm{not}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{8} \\ $$$$\:\:\Rightarrow\:\mathrm{p}=\mathrm{2}\:\mathrm{and}\:\mathrm{q}=\mathrm{4}\:\: \\ $$$$\:\:\:\:\therefore\:\mathrm{N}\:=\:\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{4}} =\:\mathrm{324} \\ $$$$ \\ $$

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

n=p^a q^b r^c ... (a+1)(b+1)(c+1)...=15=3×5 ⇒a=2, b=4 ⇒n=p^2 q^4 n is divisible by 6 ⇒n must contain 2^(≥1) 3^(≥1) ...(1) n is not divisible by 8 ⇒n may contain 2^(≤2) ...(2) from (1) and (2) we can see p=2, q=3 ⇒n=2^2 3^4 =324

$${n}={p}^{{a}} {q}^{{b}} {r}^{{c}} ... \\ $$$$\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\left({c}+\mathrm{1}\right)...=\mathrm{15}=\mathrm{3}×\mathrm{5} \\ $$$$\Rightarrow{a}=\mathrm{2},\:{b}=\mathrm{4} \\ $$$$\Rightarrow{n}={p}^{\mathrm{2}} {q}^{\mathrm{4}} \\ $$$${n}\:{is}\:{divisible}\:{by}\:\mathrm{6} \\ $$$$\Rightarrow{n}\:{must}\:{contain}\:\mathrm{2}^{\geqslant\mathrm{1}} \mathrm{3}^{\geqslant\mathrm{1}} \:\:\:\:\:...\left(\mathrm{1}\right) \\ $$$${n}\:{is}\:{not}\:{divisible}\:{by}\:\mathrm{8} \\ $$$$\Rightarrow{n}\:{may}\:{contain}\:\mathrm{2}^{\leqslant\mathrm{2}} \:\:\:...\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right)\:{we}\:{can}\:{see} \\ $$$${p}=\mathrm{2},\:{q}=\mathrm{3} \\ $$$$\Rightarrow{n}=\mathrm{2}^{\mathrm{2}} \mathrm{3}^{\mathrm{4}} =\mathrm{324} \\ $$

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