Metamath Proof Explorer


Theorem 1259lem1

Description: Lemma for 1259prm . Calculate a power mod. In decimal, we calculate 2 ^ 1 6 = 5 2 N + 6 8 == 6 8 and 2 ^ 1 7 == 6 8 x. 2 = 1 3 6 in this lemma. (Contributed by Mario Carneiro, 22-Feb-2014) (Revised by Mario Carneiro, 20-Apr-2015) (Proof shortened by AV, 16-Sep-2021)

Ref Expression
Hypothesis 1259prm.1 ⊢ N = 1259
Assertion 1259lem1 ⊢ 2 17 mod N = 136 mod N

Proof

Step Hyp Ref Expression
1 1259prm.1 ⊢ N = 1259
2 1nn0 ⊢ 1 ∈ ℕ 0
3 2nn0 ⊢ 2 ∈ ℕ 0
4 2 3 deccl ⊢ 12 ∈ ℕ 0
5 5nn0 ⊢ 5 ∈ ℕ 0
6 4 5 deccl ⊢ 125 ∈ ℕ 0
7 9nn ⊢ 9 ∈ ℕ
8 6 7 decnncl ⊢ 1259 ∈ ℕ
9 1 8 eqeltri ⊢ N ∈ ℕ
10 2nn ⊢ 2 ∈ ℕ
11 6nn0 ⊢ 6 ∈ ℕ 0
12 2 11 deccl ⊢ 16 ∈ ℕ 0
13 0z ⊢ 0 ∈ ℤ
14 8nn0 ⊢ 8 ∈ ℕ 0
15 11 14 deccl ⊢ 68 ∈ ℕ 0
16 3nn0 ⊢ 3 ∈ ℕ 0
17 2 16 deccl ⊢ 13 ∈ ℕ 0
18 17 11 deccl ⊢ 136 ∈ ℕ 0
19 5 3 deccl ⊢ 52 ∈ ℕ 0
20 19 nn0zi ⊢ 52 ∈ ℤ
21 3 14 nn0expcli ⊢ 2 8 ∈ ℕ 0
22 eqid ⊢ 2 8 mod N = 2 8 mod N
23 14 nn0cni ⊢ 8 ∈ ℂ
24 2cn ⊢ 2 ∈ ℂ
25 8t2e16 ⊢ 8 ⋅ 2 = 16
26 23 24 25 mulcomli ⊢ 2 ⋅ 8 = 16
27 9nn0 ⊢ 9 ∈ ℕ 0
28 eqid ⊢ 68 = 68
29 4nn0 ⊢ 4 ∈ ℕ 0
30 7nn0 ⊢ 7 ∈ ℕ 0
31 29 30 deccl ⊢ 47 ∈ ℕ 0
32 eqid ⊢ 125 = 125
33 0nn0 ⊢ 0 ∈ ℕ 0
34 11 dec0h ⊢ 6 = 06
35 eqid ⊢ 47 = 47
36 4cn ⊢ 4 ∈ ℂ
37 36 addlidi ⊢ 0 + 4 = 4
38 37 oveq1i ⊢ 0 + 4 + 1 = 4 + 1
39 4p1e5 ⊢ 4 + 1 = 5
40 38 39 eqtri ⊢ 0 + 4 + 1 = 5
41 7cn ⊢ 7 ∈ ℂ
42 6cn ⊢ 6 ∈ ℂ
43 7p6e13 ⊢ 7 + 6 = 13
44 41 42 43 addcomli ⊢ 6 + 7 = 13
45 33 11 29 30 34 35 40 16 44 decaddc ⊢ 6 + 47 = 53
46 3 11 deccl ⊢ 26 ∈ ℕ 0
47 eqid ⊢ 12 = 12
48 5 dec0h ⊢ 5 = 05
49 eqid ⊢ 26 = 26
50 24 addlidi ⊢ 0 + 2 = 2
51 50 oveq1i ⊢ 0 + 2 + 1 = 2 + 1
52 2p1e3 ⊢ 2 + 1 = 3
53 51 52 eqtri ⊢ 0 + 2 + 1 = 3
54 5cn ⊢ 5 ∈ ℂ
55 6p5e11 ⊢ 6 + 5 = 11
56 42 54 55 addcomli ⊢ 5 + 6 = 11
57 33 5 3 11 48 49 53 2 56 decaddc ⊢ 5 + 26 = 31
58 10nn0 ⊢ 10 ∈ ℕ 0
59 eqid ⊢ 52 = 52
60 58 nn0cni ⊢ 10 ∈ ℂ
61 3cn ⊢ 3 ∈ ℂ
62 dec10p ⊢ 10 + 3 = 13
63 60 61 62 addcomli ⊢ 3 + 10 = 13
64 54 mulridi ⊢ 5 ⋅ 1 = 5
65 1p0e1 ⊢ 1 + 0 = 1
66 64 65 oveq12i ⊢ 5 ⋅ 1 + 1 + 0 = 5 + 1
67 5p1e6 ⊢ 5 + 1 = 6
68 66 67 eqtri ⊢ 5 ⋅ 1 + 1 + 0 = 6
69 24 mulridi ⊢ 2 ⋅ 1 = 2
70 69 oveq1i ⊢ 2 ⋅ 1 + 3 = 2 + 3
71 3p2e5 ⊢ 3 + 2 = 5
72 61 24 71 addcomli ⊢ 2 + 3 = 5
73 70 72 48 3eqtri ⊢ 2 ⋅ 1 + 3 = 05
74 5 3 2 16 59 63 2 5 33 68 73 decmac ⊢ 52 ⋅ 1 + 3 + 10 = 65
75 2 dec0h ⊢ 1 = 01
76 5t2e10 ⊢ 5 ⋅ 2 = 10
77 00id ⊢ 0 + 0 = 0
78 76 77 oveq12i ⊢ 5 ⋅ 2 + 0 + 0 = 10 + 0
79 dec10p ⊢ 10 + 0 = 10
80 78 79 eqtri ⊢ 5 ⋅ 2 + 0 + 0 = 10
81 2t2e4 ⊢ 2 ⋅ 2 = 4
82 81 oveq1i ⊢ 2 ⋅ 2 + 1 = 4 + 1
83 82 39 48 3eqtri ⊢ 2 ⋅ 2 + 1 = 05
84 5 3 33 2 59 75 3 5 33 80 83 decmac ⊢ 52 ⋅ 2 + 1 = 105
85 2 3 16 2 47 57 19 5 58 74 84 decma2c ⊢ 52 ⋅ 12 + 5 + 26 = 655
86 5t5e25 ⊢ 5 ⋅ 5 = 25
87 3 5 67 86 decsuc ⊢ 5 ⋅ 5 + 1 = 26
88 54 24 76 mulcomli ⊢ 2 ⋅ 5 = 10
89 61 addlidi ⊢ 0 + 3 = 3
90 2 33 16 88 89 decaddi ⊢ 2 ⋅ 5 + 3 = 13
91 5 3 16 59 5 16 2 87 90 decrmac ⊢ 52 ⋅ 5 + 3 = 263
92 4 5 5 16 32 45 19 16 46 85 91 decma2c ⊢ 52 ⋅ 125 + 6 + 47 = 6553
93 9cn ⊢ 9 ∈ ℂ
94 9t5e45 ⊢ 9 ⋅ 5 = 45
95 93 54 94 mulcomli ⊢ 5 ⋅ 9 = 45
96 5p2e7 ⊢ 5 + 2 = 7
97 29 5 3 95 96 decaddi ⊢ 5 ⋅ 9 + 2 = 47
98 9t2e18 ⊢ 9 ⋅ 2 = 18
99 93 24 98 mulcomli ⊢ 2 ⋅ 9 = 18
100 1p1e2 ⊢ 1 + 1 = 2
101 8p8e16 ⊢ 8 + 8 = 16
102 2 14 14 99 100 11 101 decaddci ⊢ 2 ⋅ 9 + 8 = 26
103 5 3 14 59 27 11 3 97 102 decrmac ⊢ 52 ⋅ 9 + 8 = 476
104 6 27 11 14 1 28 19 11 31 92 103 decma2c ⊢ 52 ⋅ N + 68 = 65536
105 2exp16 ⊢ 2 16 = 65536
106 eqid ⊢ 2 8 = 2 8
107 eqid ⊢ 2 8 ⁢ 2 8 = 2 8 ⁢ 2 8
108 3 14 26 106 107 numexp2x ⊢ 2 16 = 2 8 ⁢ 2 8
109 104 105 108 3eqtr2i ⊢ 52 ⋅ N + 68 = 2 8 ⁢ 2 8
110 9 10 14 20 21 15 22 26 109 mod2xi ⊢ 2 16 mod N = 68 mod N
111 6p1e7 ⊢ 6 + 1 = 7
112 eqid ⊢ 16 = 16
113 2 11 111 112 decsuc ⊢ 16 + 1 = 17
114 18 nn0cni ⊢ 136 ∈ ℂ
115 114 addlidi ⊢ 0 + 136 = 136
116 9 nncni ⊢ N ∈ ℂ
117 116 mul02i ⊢ 0 ⋅ N = 0
118 117 oveq1i ⊢ 0 ⋅ N + 136 = 0 + 136
119 6t2e12 ⊢ 6 ⋅ 2 = 12
120 2 3 52 119 decsuc ⊢ 6 ⋅ 2 + 1 = 13
121 3 11 14 28 11 2 120 25 decmul1c ⊢ 68 ⋅ 2 = 136
122 115 118 121 3eqtr4i ⊢ 0 ⋅ N + 136 = 68 ⋅ 2
123 9 10 12 13 15 18 110 113 122 modxp1i ⊢ 2 17 mod N = 136 mod N