2^m −2^n = 2016 find m and n


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Question Number 175458 by Linton last updated on 30/Aug/22

$2^{m} - 2^{n} = 2016$ $f i n d m a n d n$
	Answered by Ar Brandon last updated on 30/Aug/22

	$2^{m} - 2^{n} = 2016$ $2^{n} (2^{m - n} - 1) = 2016 = 1008 \times 2 = 504 \times 4$ $= 252 \times 8 = 126 \times 16 = 63 \times 32 = (2^{6} - 1) \times (2^{5})$ $\Rightarrow n = 5, m = 6 + 5 = 11$
	Answered by Rasheed.Sindhi last updated on 01/Sep/22

	$2^{m} - 2^{n} = 2016 = 2^{5} \cdot 3^{2} . 7$ $\frac{2^{m} - 2^{n}}{2^{5}} = 3^{2} \cdot 7 = 63 = 64 - 1$ $[63 c a n b e e x p r e s s e d a s a d i f f e r e n c e$ $o f p o w e r s o f 2 i n a u n i q u e w a y]$ $2^{m - 5} - 2^{n - 5} = 2^{6} - 2^{0}$ $m - 5 = 6 \Rightarrow m = 11$ $n - 5 = 0 \Rightarrow n = 5$
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