Question

Determinati ultima cifra a nr
B=7+7 la 2+7 la 3+...+7 la 43
C=3 la 0+3 la 1+3 la 2+3 la 3+...+3 la 42
D=8+8 la 2+8 la 3+...+8 la 88
Va rog frumos sa ma ajutati nu stiu!!..

Answer 1

$\displaystyle a).7+7^2+7^3+...+7^4^3=1+2+3+...+43= \frac{43(43+1)}{2} = \\ \\ = \frac{43 \times 44}{2} = \frac{1892}{2} =946=7^9^4^6 \\ \\ U(7^9^4^6)=U(7 \times 7^9^4^5)=U(7 \times (7^5)^1^8^9)=U(7 \times 16807^1^8^9)= \\ \\ =U(7 \times 7)=U(49)=9$

$\displaystyle b).3^0+3^1+3^2+3^3+...+3^4^2 =0+1+2+3+...+42= \\ \\ = \frac{42(42+1)}{2} = \frac{42 \times 43}{2} = \frac{1806}{2} =903=3^9^0^3 \\ \\ U(3^{903})=U(3 \times 3^{902})=U(3 \times (3^2)^4^5^1)=U(3 \times 9^4^5^1)= \\ \\ =U(3 \times 9)=U(27)=7=U \big{( \frac{7-1}{2} \big )}=U \big{( \frac{6}{2} \big)}=U(3)=3$

$\displaystyle c).8+8^2+8^3+...+8^8^8=1+2+3+...+88= \frac{88(88+1)}{2} = \\ \\ = \frac{88 \times 89}{2} = \frac{7832}{2} =3916=8^{3916} \\ \\ U(8^{3916})=U(8 \times 8^{3915})=U(8 \times (8^5)^7^8^3)=U(8 \times 32768^7^8^3)= \\ \\ =U(8 \times 8)=U(64)=4$