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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-t




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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