Question

1.Fie a= $1^{33}$ + $4^{34}$ + $7^{35}$ . Aflati ultima cifra a numarului a.

2. Fie S= $2^{0}$ + $2^{1}$ + $2^{2}$ + $2^{3}$ + ... + $2^{99}$
a) Demonstrati ca S se divide cu 15
b) Aratati ca S are cel putin 30 de cifre.

Answer 1

$1.~A=1^{33}+4^{34}+7^{35}=1+4^{34}+7^{35}. \\ \\ 4^n~(n \in N^*)~are~ultima~cifra~4~daca~n~este~impar,~si~6~daca~n~este \\ par.~ \\ \\ Deci ~4^{34} ~are~ ultima ~cifra ~egala~cu~6. \\ \\ 7^m~(m \in N^*)~are~ultima~cifra~egala~cu: \\ -7,~daca~m=M_4+1 \\ -9,~daca~m=M_4+2 \\ -3,~daca~m=M_4+3 \\ -1,~daca~m=M_4 \\ \\ 35=4 \cdot 8+3 = M_4+3 \Rightarrow U_{(7^{35})} =3.$

$1+3+6=10,~deci~U_{(N)}=0. \\ \\ *U_{x} ~reprezinta~ultima~cifra~a~numarului~natural~x.$

$2.~ \\ S=2^0+2^1+2^2+2^3+...+2^{99}= \\ ~~~=(2^0+2^1+2^2+2^3)+2^4(2^0+2^1+2^2+2^3)+...+ \\ +2^{96} (2^0+2^1+2^2+2^3)= \\ ~~~=(2^0+2^1+2^2+2^3)(1+2^4+2^8+...+2^{96})= \\ ~~~=15 \cdot (1+2^4+2^8+...+2^{96}) \Rightarrow S ~\vdots~15.$

$S=1+2+2^2+2^3+...+2^{100}= \\ \\ ~~~= \frac{2^{101}-1}{2-1}= \\ \\ ~~~=2^{101}-1. \\ \\ Avem~de~demonstrat~ca~2^{101}-1~are~cel~putin~30~de~cifre . \\ \\ S\ \textgreater \ 2^{100}-1=(2^{10})^{10}-1=1024^{10}-1\ \textgreater \ 1000^{10}-1=10^{30}-1. \\ \\ S\ \textgreater \ 10^{30}-1 \Rightarrow S~are~cel~putin~30~de~cifre.$