Question

Determinati ultima cifra a numerelor:
 a)099 la puterea 51
b)11 la puterea 53+15 la puterea 53+17 la puterea 53
c)313 la puterea 100
d)68 la puterea 86
e)89 la puterea 37+ 88 la puterea 38+ 87 la puterea 39
f)71 la puterea 1000001
g)1009 la puterea 9001
h)77 la puterea 36+77 la puterea 37+77 la puterea 37+77 la puterea 39
va rog ajutati-ma <3

Answer 1

a)99^51=u(99^51)=u(9^51)=u(9^4^*12*9^3)=u(9^4)*u(9^3)=1*9=u(9)
b) u(11^53)=u(1^53)=u(1) ; u=(15^53)=u(5^53)=u(5^4^*13+5^1)=u(5^4)*u(5^1)=5*5=25=u(5); u(17^53)=u(7^53)=u(7^4^*13+7^1)=u(7^4)*u(7^1)=1*7=u(7) deci 7+1+5 =8+5=13 de unde rezulta u(3)
c) 313^100=u(3^100)==u(25^*4)=u(3^4)=u(1)
d)68^86=U(8^86)=u(8^21^*4+8^2)=u(8^4)*u(8^2)=6*4=24=u(4)
e)89^37+88^38+87^39 , deci , u(89^37)=u(9^37)=u(9^4+9^1)=u(9^4)*u(9^2)=1*9=u(9) ;
u(88^37)=u(9^38)=u(8^9^*4+9^1)=u(9^4)*u(8^2)=6*4=24=u(4) ;
u(87^39)=u(7^39)=u(7^9^*4+7^3)=u(7^4)*u(7^3)=1*3=u(3)
9+4+3=16=u(6)
f)71^10000001=u(110000001)=u(1)
g) 1009^9001=u(9^9001)=u(9^2250^*4+9^1)=u(9^4)*u(9^1)=1*9=u(9)
h)u(77^36)=u(7^36)=u(7^9^*4)=u(7^4)=u(1) ;
u(77^37)=u(7^37)=u(7^9^*4+7^1)=u(7^4)*(7^1)=1*7=u(7) ;
u(77^38)=u(7^38)=u(7^9^*4+7^2)=u(7^4)*u(7^2)=1*9=u(9) ;
u(77^39)=u(7^39)=u(7^9^*4+7^3)=u(7^4)*u(7^3)=1*3=u(3) , deci ,  1+7+9+3=10+10=20=u(0)
Ps: *= inmultire si ^ =la puterea , iar ^*=ori (dar este la putere)