Puteri ale lui 2 cu suma cifrelor neconstanta
Moderators: Laurian Filip, Filip Chindea, maky, Cosmin Pohoata
-
Alin Galatan
- Site Admin
- Posts: 247
- Joined: Tue Sep 25, 2007 9:24 pm
- Location: Bucuresti/Timisoara/Moldova Noua
Puteri ale lui 2 cu suma cifrelor neconstanta
Demonstrati ca nu exista o infinitate de puteri ale lui 2, distincte, cu aceeasi suma a cifrelor.
-
Filip Chindea
- Newton
- Posts: 324
- Joined: Thu Sep 27, 2007 9:01 pm
- Location: Bucharest
Am primit raspunsul pe MLS. Aceasta este (aproximativ) solutia postata de Gabriel D.:
Fie \( \alpha := \log_2(10) = \log(10)/\log(2) \). Definim sirul strict crescator \( a_1 = 2\alpha \), \( a_{n+1} = 1 + \lfloor \alpha a_n \rfloor \).
Evident \( a_n < C^n \), \( C > 2\alpha \) implica \( a_{n+1} \le 1 + \alpha a_n < 2\alpha C^n < C^{n+1} \). Am aratat inductiv ca \( \alpha^n \le a_n \le C^n \).
Consideram acum \( a_{k} \le n < a_{k+1} < C^{k+1} \). Deci \( j \le k \) implica \( 2^{a_j} | 2^n \). Scriem (in baza 10) \( 2^n = c_0 + c_1 \cdot 10 + \cdots \). Avem si \( 2^{a_j} | c_{a_j} \cdot 10^{a_j} + c_{a_j+1} \cdot 10^{a_j+1} + \cdots \), deci \( 2^{a_j} | c_1 + \cdots + c_{a_j - 1} \cdot 10^{a_j - 1} \). Cel putin una din cifrele \( c_{a_{j-1}}, ..., c_{a_j - 1} \) nu este zero (altfel \( 2^{a_j} \le 10^{a_{j-1}}, \alpha a_{j-1} \ge a_j > \alpha a_{j-1} \), fals). Considerand cele \( k \) astfel de "blocuri", obtinem \( k \) cifre nenule. Deci \( s(2^n) \ge k > D \log(n) - 1 \), \( D = 1/\log(C) \), o contradictie.
(Schinzel)
Fie \( \alpha := \log_2(10) = \log(10)/\log(2) \). Definim sirul strict crescator \( a_1 = 2\alpha \), \( a_{n+1} = 1 + \lfloor \alpha a_n \rfloor \).
Evident \( a_n < C^n \), \( C > 2\alpha \) implica \( a_{n+1} \le 1 + \alpha a_n < 2\alpha C^n < C^{n+1} \). Am aratat inductiv ca \( \alpha^n \le a_n \le C^n \).
Consideram acum \( a_{k} \le n < a_{k+1} < C^{k+1} \). Deci \( j \le k \) implica \( 2^{a_j} | 2^n \). Scriem (in baza 10) \( 2^n = c_0 + c_1 \cdot 10 + \cdots \). Avem si \( 2^{a_j} | c_{a_j} \cdot 10^{a_j} + c_{a_j+1} \cdot 10^{a_j+1} + \cdots \), deci \( 2^{a_j} | c_1 + \cdots + c_{a_j - 1} \cdot 10^{a_j - 1} \). Cel putin una din cifrele \( c_{a_{j-1}}, ..., c_{a_j - 1} \) nu este zero (altfel \( 2^{a_j} \le 10^{a_{j-1}}, \alpha a_{j-1} \ge a_j > \alpha a_{j-1} \), fals). Considerand cele \( k \) astfel de "blocuri", obtinem \( k \) cifre nenule. Deci \( s(2^n) \ge k > D \log(n) - 1 \), \( D = 1/\log(C) \), o contradictie.
(Schinzel)
Life is complex: it has real and imaginary components.