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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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