Question

Chứng tỏ A chia hết cho 31, biết:

A = 5 + 5 mũ 1 + 5 mũ 2 + ... + 5 mũ 101

Answer 1

\(A=5^0+5+5^2+....+5^{101}\\ =\left(5^0+5+5^2\right)+...+\left(5^{99}+5^{100}+5^{101}\right)\\ =31+...+5^{99}\left(5^0+5^1+5^2\right)\\ =31+...+31.5^{99}\\ =31\left(1+...+5^{99}\right)⋮31\\ \Rightarrow A⋮31\left(dpcm\right)\)

Answer 2

A = 5 + 5^1 + 5^2 +............+ 5^101

A = 1( 1+ 5 + 5^2)+..............+5^99(1+ 5 + 5^2)

A = 1.31 +..............+ 5^99.31

A = 31.(1 +.............+ 5^99) nên A chia hết cho 31

Answer 3

A=50+5+52+....+5101=(50+5+52)+...+(599+5100+5101)=31+...+599(50+51+52)=31+...+31.599=31(1+...+599)⋮31⇒A⋮31(dpcm)