10. With what least number must 8640 be divided
so that the quotient is a perfect cube​

1 thought on “<br />10. With what least number must 8640 be divided<br />so that the quotient is a perfect cube​”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Let us first find the prime factorization of 8640.

   [tex] \red{\rm :\longmapsto\:Prime \: factorization \: of \: 8640}[/tex]

   [tex]\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:8640 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:4320 \:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:2160\:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:1080 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:540 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:270 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:135 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:45 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:15 \:\:}} \\ {\underline{\sf{5}}}& \underline{\sf{\:\:5 \:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}[/tex]

   Hence,

   Prime factorization of 8640 is

   [tex]\bf\implies \:8640 = {2}^{3} \times {2}^{3} \times {3}^{3} \times 5[/tex]

   So, to obtain a perfect cube, 8640 must be divided by 5.

   Hence, 5 is the least number.

   So, required number is 1728.

   [tex]\rm :\longmapsto\: \sqrt[3]{1728} [/tex]

   [tex] \rm \:= \: \:\: \sqrt[3]{ {2}^{3} \times {2}^{3} \times {3}^{3} } [/tex]

   [tex] \rm \:= \: \:2 \times 2 \times 3[/tex]

   [tex] \rm \:= \: \:12[/tex]

   [tex]\bf\implies \: \sqrt[3]{1728} = 12[/tex]

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