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Question Number 8560 by FilupSmith last updated on 16/Oct/16

(n/2^k )=m, n,m,k∈Z For any integer n, how many times can you divide it by 2, such that the result is always a whole number. e.g. 120 160/2=80 80/2=40 40/2=20 20/2=10 10/2=5 5/2∉Z ∴divides 5 times (n/2^k )=m ⇒ ((160)/2^k )=5 2^k =((160)/5)=32=2^5 k=5

$$\frac{{n}}{\mathrm{2}^{{k}} }={m},\:\:\:\:\:\:\:{n},{m},{k}\in\mathbb{Z} \\ $$$$\mathrm{For}\:\mathrm{any}\:\mathrm{integer}\:{n},\:\mathrm{how}\:\mathrm{many}\:\mathrm{times} \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{divide}\:\mathrm{it}\:\mathrm{by}\:\mathrm{2},\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{result}\:\mathrm{is}\:\mathrm{always}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{number}. \\ $$$$\: \\ $$$$\mathrm{e}.\mathrm{g}.\:\mathrm{120} \\ $$$$\mathrm{160}/\mathrm{2}=\mathrm{80} \\ $$$$\mathrm{80}/\mathrm{2}=\mathrm{40} \\ $$$$\mathrm{40}/\mathrm{2}=\mathrm{20} \\ $$$$\mathrm{20}/\mathrm{2}=\mathrm{10} \\ $$$$\mathrm{10}/\mathrm{2}=\mathrm{5} \\ $$$$\mathrm{5}/\mathrm{2}\notin\mathbb{Z} \\ $$$$\therefore\mathrm{divides}\:\mathrm{5}\:\mathrm{times} \\ $$$$\frac{{n}}{\mathrm{2}^{{k}} }={m}\:\:\:\Rightarrow\:\:\frac{\mathrm{160}}{\mathrm{2}^{{k}} }=\mathrm{5} \\ $$$$\mathrm{2}^{{k}} =\frac{\mathrm{160}}{\mathrm{5}}=\mathrm{32}=\mathrm{2}^{\mathrm{5}} \\ $$$${k}=\mathrm{5} \\ $$

Commented by 123456 last updated on 16/Oct/16

(n/2^k )=m (n,m,k)∈N 2^k =(n/m) k=lg((n/m)) max k→min m

$$\frac{{n}}{\mathrm{2}^{{k}} }={m}\:\:\left({n},{m},{k}\right)\in\mathbb{N} \\ $$$$\mathrm{2}^{{k}} =\frac{{n}}{{m}} \\ $$$${k}=\mathrm{lg}\left(\frac{{n}}{{m}}\right) \\ $$$$\mathrm{max}\:{k}\rightarrow\mathrm{min}\:{m} \\ $$

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