Metamath Proof Explorer


Theorem 2503lem1

Description: Lemma for 2503prm . Calculate a power mod. In decimal, we calculate 2 ^ 1 8 = 5 1 2 ^ 2 = 1 0 4 N + 1 8 3 2 == 1 8 3 2 . (Contributed by Mario Carneiro, 3-Mar-2014) (Revised by Mario Carneiro, 20-Apr-2015) (Proof shortened by AV, 16-Sep-2021)

Ref Expression
Hypothesis 2503prm.1 N=2503
Assertion 2503lem1 218modN=1832modN

Proof

Step Hyp Ref Expression
1 2503prm.1 N=2503
2 2nn0 20
3 5nn0 50
4 2 3 deccl 250
5 0nn0 00
6 4 5 deccl 2500
7 3nn 3
8 6 7 decnncl 2503
9 1 8 eqeltri N
10 2nn 2
11 9nn0 90
12 10nn0 100
13 4nn0 40
14 12 13 deccl 1040
15 14 nn0zi 104
16 1nn0 10
17 3 16 deccl 510
18 17 2 deccl 5120
19 8nn0 80
20 16 19 deccl 180
21 3nn0 30
22 20 21 deccl 1830
23 22 2 deccl 18320
24 8p1e9 8+1=9
25 6nn0 60
26 2exp8 28=256
27 eqid 25=25
28 16 dec0h 1=01
29 2t2e4 22=4
30 ax-1cn 1
31 30 addlidi 0+1=1
32 29 31 oveq12i 22+0+1=4+1
33 4p1e5 4+1=5
34 32 33 eqtri 22+0+1=5
35 5t2e10 52=10
36 16 5 31 35 decsuc 52+1=11
37 2 3 5 16 27 28 2 16 16 34 36 decmac 252+1=51
38 6t2e12 62=12
39 2 4 25 26 2 16 37 38 decmul1c 282=512
40 2 19 24 39 numexpp1 29=512
41 40 oveq1i 29modN=512modN
42 9cn 9
43 2cn 2
44 9t2e18 92=18
45 42 43 44 mulcomli 29=18
46 eqid 1832=1832
47 21 16 deccl 310
48 2 16 deccl 210
49 eqid 250=250
50 eqid 183=183
51 eqid 31=31
52 eqid 18=18
53 1p1e2 1+1=2
54 8p3e11 8+3=11
55 16 19 21 52 53 16 54 decaddci 18+3=21
56 3p1e4 3+1=4
57 20 21 21 16 50 51 55 56 decadd 183+31=214
58 48 nn0cni 21
59 58 addridi 21+0=21
60 3 2 deccl 520
61 eqid 104=104
62 60 nn0cni 52
63 eqid 52=52
64 2p2e4 2+2=4
65 3 2 2 63 64 decaddi 52+2=54
66 62 43 65 addcomli 2+52=54
67 2 dec0u 102=20
68 5p1e6 5+1=6
69 67 68 oveq12i 102+5+1=20+6
70 eqid 20=20
71 6cn 6
72 71 addlidi 0+6=6
73 2 5 25 70 72 decaddi 20+6=26
74 69 73 eqtri 102+5+1=26
75 4t2e8 42=8
76 75 oveq1i 42+4=8+4
77 8p4e12 8+4=12
78 76 77 eqtri 42+4=12
79 12 13 3 13 61 66 2 2 16 74 78 decmac 1042+2+52=262
80 3 dec0u 105=50
81 43 addlidi 0+2=2
82 80 81 oveq12i 105+0+2=50+2
83 eqid 50=50
84 3 5 2 83 81 decaddi 50+2=52
85 82 84 eqtri 105+0+2=52
86 5cn 5
87 4cn 4
88 5t4e20 54=20
89 86 87 88 mulcomli 45=20
90 2 5 31 89 decsuc 45+1=21
91 12 13 5 16 61 28 3 16 2 85 90 decmac 1045+1=521
92 2 3 2 16 27 59 14 16 60 79 91 decma2c 10425+21+0=2621
93 14 nn0cni 104
94 93 mul01i 1040=0
95 94 oveq1i 1040+4=0+4
96 87 addlidi 0+4=4
97 13 dec0h 4=04
98 95 96 97 3eqtri 1040+4=04
99 4 5 48 13 49 57 14 13 5 92 98 decma2c 104250+183+31=26214
100 eqid 10=10
101 3cn 3
102 101 mullidi 13=3
103 00id 0+0=0
104 102 103 oveq12i 13+0+0=3+0
105 101 addridi 3+0=3
106 104 105 eqtri 13+0+0=3
107 101 mul02i 03=0
108 107 oveq1i 03+1=0+1
109 108 31 28 3eqtri 03+1=01
110 16 5 5 16 100 28 21 16 5 106 109 decmac 103+1=31
111 4t3e12 43=12
112 16 2 2 111 64 decaddi 43+2=14
113 12 13 2 61 21 13 16 110 112 decrmac 1043+2=314
114 6 21 22 2 1 46 14 13 47 99 113 decma2c 104 N+1832=262144
115 eqid 512=512
116 12 2 deccl 1020
117 eqid 51=51
118 eqid 102=102
119 86 30 68 addcomli 1+5=6
120 16 5 3 16 100 117 119 31 decadd 10+51=61
121 7nn0 70
122 6p1e7 6+1=7
123 121 dec0h 7=07
124 122 123 eqtri 6+1=07
125 31 oveq2i 55+0+1=55+1
126 5t5e25 55=25
127 2 3 68 126 decsuc 55+1=26
128 125 127 eqtri 55+0+1=26
129 86 mullidi 15=5
130 129 oveq1i 15+7=5+7
131 7cn 7
132 7p5e12 7+5=12
133 131 86 132 addcomli 5+7=12
134 130 133 eqtri 15+7=12
135 3 16 5 121 117 124 3 2 16 128 134 decmac 515+6+1=262
136 86 43 35 mulcomli 25=10
137 16 5 31 136 decsuc 25+1=11
138 17 2 25 16 115 120 3 16 16 135 137 decmac 5125+10+51=2621
139 17 nn0cni 51
140 139 mulridi 511=51
141 43 mulridi 21=2
142 141 oveq1i 21+2=2+2
143 142 64 eqtri 21+2=4
144 17 2 2 115 16 140 143 decrmanc 5121+2=514
145 3 16 12 2 117 118 18 13 17 138 144 decma2c 51251+102=26214
146 43 mullidi 12=2
147 2 3 16 117 35 146 decmul1 512=102
148 2 17 2 115 147 29 decmul1 5122=1024
149 18 17 2 115 13 116 145 148 decmul2c 512512=262144
150 114 149 eqtr4i 104 N+1832=512512
151 9 10 11 15 18 23 41 45 150 mod2xi 218modN=1832modN