Question

Cati divizori proprii are nr : (111111+222222+..999999):(111+222+...+999) ? 100 PUNCTE

Answer 1

$Avem:~ \overline{ \underbrace{aaa} \underbrace{aaa}}=1000 \cdot \overline{aaa}+ \overline{aaa}=1001 \cdot \overline{aaa}.$

$Prima~paranteza~este~egala~cu: \\ \\ 1001 \cdot 111 + 1001 \cdot 222 + 1001 \cdot 333 +... + 1001 \cdot 999= \\ \\ =1001(111+222+333+...+999). \\ \\ Notez~111+222+333+...+999=t~(se~poate~calcula,~dar \\ \\ nu~este~necesar!)\\ \\ Deci~numarul~din~enunt~este~egal~cu: \\ \\ 1001t:t=1001.\\ \\ Si~pentru~a~calcula~numarul~de~divizori~naturali~ai~lui \\ \\ 1001,~folosim ~urmatoarea~proprietate:$

$Daca~x=p_1^{r_1} \cdot p_2^{r_2} \cdot ... \cdot p_n^{r_n},~atunci~numarul~x~are \\ \\ (r_1+1)(r_2+1)...(r_n+1)~divizori~naturali.~(p_1,~p_2...p_n \\ \\ fiind~numerele~prime~din~descompunerea~lui~x~in~produs \\ \\ de~factori~primi;~x \in N- \{0;1 \}) \\ \\ 1001=7^1 \cdot 11^1 \cdot 13^1 \Rightarrow (1+1)(1+1)(1+1)~divizori,~din~care \\ \\ proprii~sunt~in~numar~de~6~(se~exclud~1~si~1001).$

$D_{1001}= \{1, \underbrace{7,11,13,77,91,143},1001 \}.$