Question

Se dau nr: a=(25^n:5^n+3^4-17-5^n)x[(2^5-5^2-7):1993]+1992,nN b=32x1992^5-3984^5+2^3x3x83 c=10^3+9x10^2+9x10+1 CALCULATI: (2a-b-c)^1907

Answer 1

a=(25^n:5^n+3^4-17-5^n)x[(2^5-5^2-7):1993]+1992=
=(5^n+81-17-5^n)x[(32-25-7):1993]+1992=
=81 x(0:1993)+1992=0+1992=1992

b=32x1992^5-3984^5+2^3x3x83 =
=32x1992^5-2^5 x 1992^5+2^3x3x83 =
=1992^5(32-32)+8x3x83 =
=0+1992=1992

c=10^3+9x10^2+9x10+1 =
=10(10^2+9x10+9)+1 =
=10(100+90+9)+1 =
=10x199+1 =
=1990+1=1991

 (2a-b-c)^1907=(2x1992-1992-1991)^1907=(3981-3983)^1907=1^1907=1