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Question Number 18949 by Tinkutara last updated on 01/Aug/17

Find the number of numbers ≤ 10^8 which are neither perfect squares, nor perfect cubes, nor perfect fifth powers.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{numbers}\:\leqslant\:\mathrm{10}^{\mathrm{8}} \\ $$$$\mathrm{which}\:\mathrm{are}\:\mathrm{neither}\:\mathrm{perfect}\:\mathrm{squares},\:\mathrm{nor} \\ $$$$\mathrm{perfect}\:\mathrm{cubes},\:\mathrm{nor}\:\mathrm{perfect}\:\mathrm{fifth}\:\mathrm{powers}. \\ $$

Commented by prakash jain last updated on 03/Aug/17

i retyped the answer.

$$\mathrm{i}\:\mathrm{retyped}\:\mathrm{the}\:\mathrm{answer}. \\ $$

Answered by prakash jain last updated on 03/Aug/17

no of perfect squares=10^4 no of perfect cubes=464 no of perfect power 5=39 no of perfect power 6=21 no of perfect power 15=3 no of perfect power 10=6 no of perfect power 30=1 number of perfect squares cubes and power 5 n=10000+464+39−21−3−6+1 requited answer=10^8 −n

$$\mathrm{no}\:\mathrm{of}\:\mathrm{perfect}\:\mathrm{squares}=\mathrm{10}^{\mathrm{4}} \\ $$$$\mathrm{no}\:\mathrm{of}\:\mathrm{perfect}\:\mathrm{cubes}=\mathrm{464} \\ $$$$\mathrm{no}\:\mathrm{of}\:\mathrm{perfect}\:\mathrm{power}\:\mathrm{5}=\mathrm{39} \\ $$$$\mathrm{no}\:\mathrm{of}\:\mathrm{perfect}\:\mathrm{power}\:\mathrm{6}=\mathrm{21} \\ $$$$\mathrm{no}\:\mathrm{of}\:\mathrm{perfect}\:\mathrm{power}\:\mathrm{15}=\mathrm{3} \\ $$$$\mathrm{no}\:\mathrm{of}\:\mathrm{perfect}\:\mathrm{power}\:\mathrm{10}=\mathrm{6} \\ $$$$\mathrm{no}\:\mathrm{of}\:\mathrm{perfect}\:\mathrm{power}\:\mathrm{30}=\mathrm{1} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{perfect}\:\mathrm{squares} \\ $$$$\mathrm{cubes}\:\mathrm{and}\:\mathrm{power}\:\mathrm{5} \\ $$$$\mathrm{n}=\mathrm{10000}+\mathrm{464}+\mathrm{39}−\mathrm{21}−\mathrm{3}−\mathrm{6}+\mathrm{1} \\ $$$$\mathrm{requited}\:\mathrm{answer}=\mathrm{10}^{\mathrm{8}} −\mathrm{n} \\ $$

Commented by Tinkutara last updated on 03/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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