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Question Number 118181 by mathocean1 last updated on 15/Oct/20

find all numbers >1 from N which  their cube are <18360

$${find}\:{all}\:{numbers}\:>\mathrm{1}\:{from}\:\mathbb{N}\:{which} \\ $$ $${their}\:{cube}\:{are}\:<\mathrm{18360} \\ $$

Commented bymathocean1 last updated on 15/Oct/20

thanks

$${thanks} \\ $$

Answered by TANMAY PANACEA last updated on 15/Oct/20

18360=3×6120               =3×3×2040               =3^3 ×680               =3^3 ×2^3 ×85  3^3 ×2^3 ×4^3 <3^3 ×2^3 ×85<3^3 ×2^3 ×5^3   so the numbers are  2^3 =8  3^3 =27  4^3 =64  2^3 ×3^3 =216  2^3 ×4^3 =512  3^3 ×4^3 =144×12=1728  2^3 ×3^3 ×4^3 =13824  so required answer  {2,3,4,6,8,12,24}

$$\mathrm{18360}=\mathrm{3}×\mathrm{6120} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}×\mathrm{3}×\mathrm{2040} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}^{\mathrm{3}} ×\mathrm{680} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{3}} ×\mathrm{85} \\ $$ $$\mathrm{3}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{3}} ×\mathrm{4}^{\mathrm{3}} <\mathrm{3}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{3}} ×\mathrm{85}<\mathrm{3}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{3}} ×\mathrm{5}^{\mathrm{3}} \\ $$ $${so}\:{the}\:{numbers}\:{are} \\ $$ $$\mathrm{2}^{\mathrm{3}} =\mathrm{8} \\ $$ $$\mathrm{3}^{\mathrm{3}} =\mathrm{27} \\ $$ $$\mathrm{4}^{\mathrm{3}} =\mathrm{64} \\ $$ $$\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{3}} =\mathrm{216} \\ $$ $$\mathrm{2}^{\mathrm{3}} ×\mathrm{4}^{\mathrm{3}} =\mathrm{512} \\ $$ $$\mathrm{3}^{\mathrm{3}} ×\mathrm{4}^{\mathrm{3}} =\mathrm{144}×\mathrm{12}=\mathrm{1728} \\ $$ $$\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{3}} ×\mathrm{4}^{\mathrm{3}} =\mathrm{13824} \\ $$ $${so}\:{required}\:{answer} \\ $$ $$\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{6},\mathrm{8},\mathrm{12},\mathrm{24}\right\} \\ $$

Commented bymathocean1 last updated on 15/Oct/20

thanks

$${thanks} \\ $$

Commented bymr W last updated on 15/Oct/20

27^3 =19683>18360   26^3 =17576<18360 ✓  ⇒2≤n≤26

$$\mathrm{27}^{\mathrm{3}} =\mathrm{19683}>\mathrm{18360}\: \\ $$ $$\mathrm{26}^{\mathrm{3}} =\mathrm{17576}<\mathrm{18360}\:\checkmark \\ $$ $$\Rightarrow\mathrm{2}\leqslant{n}\leqslant\mathrm{26} \\ $$

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