Question

DETERMINAȚI CATUL SI RESTUL ÎMPĂRȚIRII NUMĂRULUI a = 1+2+2 la a 2 a +2 la a 3a + 2 la a 4a...+2 la 2014 la 15​

Answer 1

$2^0+2^1+2^2+2^3 = 15\\ \\ \\1+2+2^2+2^3+2^4+...+2^{2014} =\\ \\ = (2^{2014}+2^{2013}+2^{2012}+2^{2011})+(2^{2010}+2^{2009}+2^{2008}+2^{2007})+...+\\+...\\ \\ \text{Avem 2015 termeni:}\\ 2015:4 = 503 \text{ rest }3 \\ \\ = (2^{2014}+2^{2013}+2^{2012}+2^{2011})+(2^{2010}+2^{2009}+2^{2008}+2^{2007})+...+\\+(2^6+2^5+2^4+2^3)+2^2+2^1+2^0 = \\ \\ = 2^{2011}(2^3+2^2+2^1+2^0)+2^{2007}(2^3+2^2+2^1+2^0)+...+\\ +2^3(2^3+2^2+2^1+2^0)+2^2+2^1+2^0 =$

$= 15\cdot \Big(2^{2011}+2^{2007}+...+2^{3}\Big)+2^2+2^1+2^0 = \\ \\ = 15\cdot \Big(2^3+2^7+2^{11}+...+2^{2007}+2^{2011}\Big)+4+2+1 = \\ \\ =15\cdot \Big(2^3+2^7+2^{11}+...+2^{2007}+2^{2011}\Big)+7 \\ \\ \Rightarrow \text{Catul este }2^3+2^7+2^{11}+...+2^{2007}+2^{2011}\\ \Rightarrow \text{Restul este }7$