Question

Se dau numerele n=1+3+5+7+...+2017 si m=2+4+6+8+...+2016.Calculati restul impartirii lui n la m

Answer 1

1+3+5+...+2k-1=k^2

2k-1=2017
2k=2018
k=1009

1+3+5+...+2017=1009^2

1+2+3+...+a=ax(a+1)/2

2+4+6+...+2016=2(1+2+3+...+1008)=2(1008x1009)/2=1008x1009

n/m=(1009^2)/(1008x1009)=1009/1008=1 si rest 1

1009=1x1008+1

Answer 2

n= 1+3+5+...+2017
n= (2017:2)x 2018:2
n=1008,5x2018:2
n= 2034144,5:2
n=1017072,25

m=2+4+6+...+2016
m=2x(1+2+3+...+1008)
m=2x 1008x 1009:2
m= 1008x1009
m= 1017072

=> 1017072,25: 1017072= 1 rest 25
=> rest=25