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


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

$f i n d a l l n u m b e r s > 1 f r o m N w h i c h$ $t h e i r c u b e a r e < 18360$
Commented by mathocean1 last updated on 15/Oct/20

$t h a n k s$
	Answered by TANMAY PANACEA last updated on 15/Oct/20
	$18360=3×6120 =3×3×2040 =3^3 ×680 =3^3 ×2^3 ×85 3^3 ×2^3 ×4^3 <3^3 ×2^3 ×85<3^3 ×2^3 ×5^3 so the numbers are 2^3 =8 3^3 =27 4^3 =64 2^3 ×3^3 =216 2^3 ×4^3 =512 3^3 ×4^3 =144×12=1728 2^3 ×3^3 ×4^3 =13824 so required answer {2,3,4,6,8,12,24}$
	$18360 = 3 \times 6120$ $= 3 \times 3 \times 2040$ $= 3^{3} \times 680$ $= 3^{3} \times 2^{3} \times 85$ $3^{3} \times 2^{3} \times 4^{3} < 3^{3} \times 2^{3} \times 85 < 3^{3} \times 2^{3} \times 5^{3}$ $s o t h e n u m b e r s a r e$ $2^{3} = 8$ $3^{3} = 27$ $4^{3} = 64$ $2^{3} \times 3^{3} = 216$ $2^{3} \times 4^{3} = 512$ $3^{3} \times 4^{3} = 144 \times 12 = 1728$ $2^{3} \times 3^{3} \times 4^{3} = 13824$ $s o r e q u i r e d a n s w e r$ ${2, 3, 4, 6, 8, 12, 24}$
	Commented by mathocean1 last updated on 15/Oct/20

	$t h a n k s$
	Commented by mr W last updated on 15/Oct/20

	$27^{3} = 19683 > 18360$ $26^{3} = 17576 < 18360 ✓$ $\Rightarrow 2 ⩽ n ⩽ 26$
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