Question

Exercitiul A15 , va rog!

Answer 1

$\displaystyle Din~enunt~avem~a_1=2,~a_2=5,~a_3=1,~a_4=4,~a_5=3. \\ \\ Pentru~a~calcula~suma,~trebuie~sa~stim~ca~atunci~cand~\sigma \\ \\ parcurge~S_5,~numarul~\sigma(i)~va~fi~egal~pe~rand~cu~1,2,3,4,5, \\ \\ de~acelasi~numar~de~ori.~(Pentru~i= \overline{1,5}.) \\ \\ Intr-adevar,~pentru~permutarile~cu~proprietatea~\sigma(1)=1, \\ \\ mai~ramane~sa~dam~valorile~2,3,4,5~numerelor~\sigma(1), \sigma(2), \sigma(3), \sigma(4) \\ \\ in~toate~modurile~posibile.~Acest~lucru~se~poate~face~in~4!=24 \\ \\ moduri.$

$\displaystyle Analog,\sigma(1)=2~se~intampla~de~24~de~ori; \\ \\ \sigma(1)=3 ~se~intampla~de~24~de~ori; \\ \\ \sigma(1)=4~se~intampla~de~24~de~ori; \\ \\ \sigma(1)=5~se~intampla~de~24~de~ori. \\ \\ Acelasi~lucru~pentru~\sigma(2),\sigma(3), \sigma(4), \sigma(5).$

$\displaystyle Acum,~daca~folosim~descompunerea~in~baza~10,~adica \\ \\ \overline{ABCDE}=10^4A+10^3B+10^2C+10D+E,~si~ne~gandim~ca \\ \\  \sigma(1)~va~fi~de~24~de~ori~egal~cu~1,~cu~2,~cu~3,~cu~4,~cu~5,~va \\ \\ rezulta~ca ~a_{\sigma(1)}~va~fi~de~24~de~ori~egal~cu~2,~cu~5,~cu~1,~cu~4,~cu~3. \\ \\ Deci~pe~pozitia~A~apar~cifrele~1,2,3,4,5~fiecare~de~cate~24~de~ori. \\ \\ (la~fel~pentru~pozitiile~B,C,D,E).$

$\displaystyle Deci~suma~respectiva~este \\ \\ 24 \cdot (1+2+3+4+5) \cdot 10^4+24 \cdot (1+2+3+4+5) \cdot 10^3+ \\ \\ + 24 \cdot (1+2+3+4+5) \cdot 10^2+24(1+2+3+4+5) \cdot 10+ \\ \\+24 \cdot (1+2+3+4+5)= \\ \\ =360 \cdot 10^4+360 \cdot 10^3+360 \cdot 10^2+ 360 \cdot 10+ 360= \\ \\ =360 \cdot 11111= \\ \\ = 3999960.$