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16th KVJ Solutions9

Here you find the solutions of  16th KVJ Solutions9 in detail explanation. 

9Q) Find the number of numbers \( \leq 10^{8} \) which are neither perfect squares, nor perfect cubes, nor perfect fifth powers .

Solution:

  Let us find number of numbers which are perfect squares , perfect cubes and perfect fifth powers \( \leq 10^{8}\).

 i) Finding number of perfect squares  \( \leq 10^{8}\) .

    \( \sqrt{10^{8}} =10^{4}=10000\)

   Therefore number of perfect squares \( \leq 10^{8}\) are 10000 .

 ii) Finding number of perfect cubes \( \leq 10^{8}\).

   Let y =\(\sqrt[3]{10^{8}}\)

         log(y) = \(\frac{8}{3} *log(10)\)

         log(y) =2.6667

             y= antilog (2.6667)

             y=4.641*10^2=464.1

  We need integers so 464 perfect cubes \( \leq 10^{8}\).

 Therefore number of perfect cubes\( \leq 10^{8}\) = 464

 iii) Finding number of perfect fifth powers.

   Let z= \(\sqrt[5]{10^{8}}\)

  log(z) = \(\frac{8}{5} *log(10)\)

  log(z) =8/5 =1.6000

     z = antilog (1.6000)=3.981x10=39.81

   Therefore number of perfect fifth powers \( \leq 10^{8}\) =39.

 Hence total number of numbers which are either perfect squares , perfect cubes or perfect fifth powers = 10000+464+39=10503

Therefore the number of numbers \( \leq 10^{8} \) which are neither perfect squares, nor perfect cubes, nor perfect fifth powers =10^8 -10503 =9,99,89,497.

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This solution is based on logarithmic tables as calculators are not allowed in examination hall.

Please review  once16th KVJ Solutions9 and click NEXT link to get solution for 10th question.

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