Question Number 175458 by Linton last updated on 30/Aug/22
$$\mathrm{2}^{{m}} −\mathrm{2}^{{n}} =\:\mathrm{2016} \\ $$$${find}\:{m}\:{and}\:{n} \\ $$
Answered by Ar Brandon last updated on 30/Aug/22
$$\mathrm{2}^{{m}} −\mathrm{2}^{{n}} =\mathrm{2016} \\ $$$$\mathrm{2}^{{n}} \left(\mathrm{2}^{{m}−{n}} −\mathrm{1}\right)=\mathrm{2016}=\mathrm{1008}×\mathrm{2}=\mathrm{504}×\mathrm{4} \\ $$$$=\mathrm{252}×\mathrm{8}=\mathrm{126}×\mathrm{16}=\mathrm{63}×\mathrm{32}\:=\left(\mathrm{2}^{\mathrm{6}} −\mathrm{1}\right)×\left(\mathrm{2}^{\mathrm{5}} \right) \\ $$$$\Rightarrow{n}=\mathrm{5},\:{m}=\mathrm{6}+\mathrm{5}=\mathrm{11} \\ $$
Answered by Rasheed.Sindhi last updated on 01/Sep/22
![2^m −2^n =2016=2^5 ∙3^2 .7 ((2^m −2^n )/2^5 )=3^2 ∙7=63=64−1 [63 can be expressed as a difference of powers of 2 in a unique way] 2^(m−5) −2^(n−5) =2^6 −2^0 m−5=6⇒m=11 n−5=0⇒n=5](https://www.tinkutara.com/question/Q175520.png)
$$\mathrm{2}^{{m}} −\mathrm{2}^{{n}} =\mathrm{2016}=\mathrm{2}^{\mathrm{5}} \centerdot\mathrm{3}^{\mathrm{2}} .\mathrm{7} \\ $$$$\frac{\mathrm{2}^{{m}} −\mathrm{2}^{{n}} }{\mathrm{2}^{\mathrm{5}} }=\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{7}=\mathrm{63}=\mathrm{64}−\mathrm{1} \\ $$$$\left[\mathrm{63}\:{can}\:{be}\:{expressed}\:{as}\:{a}\:{difference}\right. \\ $$$$\left.{of}\:{powers}\:{of}\:\mathrm{2}\:{in}\:{a}\:{unique}\:{way}\right] \\ $$$$\mathrm{2}^{{m}−\mathrm{5}} −\mathrm{2}^{{n}−\mathrm{5}} =\mathrm{2}^{\mathrm{6}} −\mathrm{2}^{\mathrm{0}} \\ $$$${m}−\mathrm{5}=\mathrm{6}\Rightarrow{m}=\mathrm{11} \\ $$$${n}−\mathrm{5}=\mathrm{0}\Rightarrow{n}=\mathrm{5} \\ $$