ملف الرياضيات السنة الثالثة ثانوي تمرين شامل الحل حول محور القسمة في زاد للأستاذ قويسم الخليل

54 ≡2[7] 55 ≡3[7] 56 ≡1[7] 5n≡ n= 6k 6k+ 1 6k+ 2 6k+ 3 6k+ 4 6k+ 5 [7] 52023 20231444 202120232025 512k+6 518k+1 401962 1962505 152k+1 52023 52023 ≡53(337)+1[7] 52023 ≡5[7] 20231444 2023 ≡0[7] 20231444 ≡01444[7] 20231444 ≡0[7] 202120232025 2023 ≡1[6] 20232025 ≡12025[6] 20232025 = 6k+ 1 2021 ≡5[7] 202120232025 ≡520232025[7] 202120232025 ≡56k+1[7] 202120232025 ≡5[7] 512k+6 512k+6 ≡56(2k+1)[7] 512k+6 ≡56k′[7] k′ ∈N 512k+6 ≡1[7] 1918k+4 1918k+4 ≡56(3k)+4[7] 1918k+4 ≡56k′+4[7] k′ ∈N 1918k+4 ≡2[7] 401962 401962 ≡56(327)[7] 401962 ≡56k[7] 401962 ≡1[7] 1962509 1962509 ≡26(84)+5[7] 1962509 ≡(−5)6k+5[7] 1962509 ≡(−1 × 5)6k+5[7] 1962509 ≡(−1)6k+5 × (5)6k+5[7] 1962509 ≡−1 × 3[7] 1962509 ≡4[7] 1252k+1 1252k+1 ≡62k+1[7] 1252k+1 ≡(−1)2k+1[7] 1252k+1 ≡−1[7] (2k+ 1) 1252k+1 ≡6[7] An An≡4 × 56n+3 + 52966 + 20222n+1 + 1[7] An≡4 × 56n+3 + 56(294)+2 + 62n+1 + 1[7] An≡4 × 6 + 4 + (−1)2n+1 + 1[7] An≡4 × 6 + 4 −1 + 1[7] An≡28[7] An≡0[7] An n∈N Bn≡56n+2 × (4n−1)[7] Bn≡5n× 56n+1 −56n+2 + 3n× 196n+2[7] Bn≡n× 56n+1+1 −56n+2 + 3n× 56n+2[7] Bn≡n× 56n+2 −56n+2 + 3n× 56n+2[7] Bn≡56n+2(4n−1)[7] Bn Bn≡0[7] 56n+2(4n−1) ≡0[7] 4(4n−1) ≡0[7] { 4n−1 ≡0[7] 4 ≡0[7] 4n−1 ≡0[7] 4n≡1[7] 8n≡2[7] n≡2[7] n= 7k+ 2 II {5n+ 56n+2 −56n+ 3n−2 ≡0[7] 5n≡4[6] −n≡4[6] n≡−4[6] n≡2[6] n= 6α+ 2 α∈N 5n+ 56n+2 −56n+ 3n−2 ≡0[7] 56α+2 + 56n+2 −56n+ 3n−2 ≡0[7] 4 + 4 −1 + 3n−2 ≡0[7] 3n+ 5 ≡0[7] 3n≡−5[7] 3n≡2[7] −2 × 3n≡−2 × 2[7] −6n≡−4[7] n≡3[7] n= 7β+ 3 β∈N {5n+ 56n+2 −56n+ 3n−2 ≡0[7] 5n≡4[6] {n= 6α+ 2 n= 7β+ 3 6α+ 2 = 7β+ 3 6α= 7β+ 1 6α≡1[7] −α≡1[7] α≡6[7] α= 7k′ + 6 k′ ∈N α n= 6α+ 2 n= 6(7k′ + 6) + 2 n= 42k′ + 38 5n+ 6n+ 7n+ 8n+ 9n≡0[7] 5n+ (−1)n+ 0n+ 1n+ 2n≡0[7] 5n+ (−1)n+ 1 + (−5)n≡0[7] 5n+ (−1)n+ 1 + (−1)n(5)n≡0[7] 5n+ (−1)n+ (−1)n(5)n≡6[7] 5n+ 1 + 5n≡6[7] 2 × 5n≡5[7] −6 × 5n≡−15[7] 5n≡6[7] n= 6p+ 3 5n+ (−1)n+ (−1)n(5)n≡6[7] 5n−1 −5n≡6[7] −1 ≡6[7] 6 ≡6[7] n= 2k+ 1 5n+ 6n+ 7n+ 8n+ 9n≡0[7] n= 2k+ 1 III 56n+ n2 −2n−1 ≡0[7] 1 + n2 −2n−1 ≡0[7] n2 −2n≡0[7] n(n−2) ≡0[7] [7] n≡ [7] n−2 ≡ [7] n(n−2) ≡ 56n+ n2 −2n−1 ≡0[7] n∈{(7k); (7k+ 2): k∈N} 2 × 56n+2 −56n+ 2n2 −n−1 ≡0[7] 2 × 4 −1 + 2n2 −n−1 ≡0[7] 2n2 −n−1 ≡0[7] 2 (n+ 2) (n−1) ≡0[7] (2n+ 1)(n−1) ≡0[7] [7] n≡ [7] n−1 ≡ [7] 2n+ 1 ≡ [7] (2n+ 1)(n−1) ≡ 2 × 56n+2 −56n+ 2n2 −n−1 ≡0[7] n∈{(7k+ 1); (7k+ 3): k∈N} {N= 1n30201 N≡0[7] N= 1n30201 1 + 2(5)2 + 3(5)4 + n(5)5 + (5)6 N≡0[7] 1 + 2(5)2 + 3(5)4 + n(5)5 + (5)6 ≡0[7] 1 + 2 × 4 + 3 × 2 + 3n+ 1 ≡0[7] 3n≡−2[7] −2 × 3n≡−2 × 5[7] −6n≡−10[7] n≡4[7] n= 7k+ 4 IV x 5x≡3[7] 3 × 5x≡3 × 3[7] 15x≡9[7] x≡2[7] x= 7k+ 2 k∈Z 1075 1505 1505 = 1075(1) + 430 1075 = 430(2) + 215 430 = 215(2) + 0 PGCD(1075; 1505) = 215 (E) Z2 | (E) Z2 (x0; y0) (E) 215 = 1075 −430(2) 215 = 1075 −(1505 −1075)(2) 215 = 1075(3) −1505(2) 215 × 3 = 1075(3 × 3) −1505(2 × 3) 645 = 1075(9) −1505(6) (9; 6) (E) (E): 1075x−1505y= 645 (E′): 5x−7y= 3