Question

Să se arate că nr 7•32^37-7^14 este divizibil cu 5

Answer 1

$(a+b)^n = M_a+b^n \\ \\ M_a\text{ inseamna multiplu de a}$

$7\cdot 32^{37}-7^{14} =7\cdot (25+7)^{37}-7^{14} = \\ \\ = 7\cdot (M_5+7^{37})-7^{14} = M_5+7^{38}-7^{14} = \\ \\=M_5+49^{19}-49^{7}= M_5 +(50-1)^{19}-(50-1)^{7} = \\ \\ = M_{5}+\Big[M_{5}+(-1)^{19}\Big]-\Big[M_{5}+(-1)^{7}\Big] = \\ \\ = M_5+M_{5}-1-(M_5-1) = \\ \\ = M_5-1+1 = \\ \\ = M_5 \\ \\ \Rightarrow 7\cdot 32^{37}-7^{14}\,\,\vdots \,\,5$