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TRANSCRIPT
100
101
102
2× 102 + 5× 101 + 3× 100 = 253
2n−1
2n
1× 20 = 11× 21 = 20× 22 = 01× 23 = 8
= 11
bt− x = 1 +
∑
t−1i=0 (b− 1)× bi
−∑
t−1i=0 ai × bi = 1 +
∑
t−1i=0 (b− 1− ai)× bi
−4 = 100−3 = 101−2 = 110−1 = 111
0 = 0001 = 0012 = 0103 = 011
x = m× be, e ∈ N
5, 5× 103 5, 5 3
2× 82 + 1× 81 + 0× 80 = 21053 = 65 = 6× 81 + 5× 80
53 = 77 = 7× 81 + 7× 80
136 = 210 = 2× 82 + 1× 81 + 0× 80
8× 161 + 8× 160 = 88
16 = 2× 2× 2× 2
53 = 3563 = 3F
60/2 = 30 030/2 = 15 015/2 = 7 17/2 = 3 13/2 = 1 11/2 = 0 1
60/8 = 7 47/8 = 0 7
→
∆
×
(
27×6 + 26×6+1 + 25×6+2 + · · ·)
010 1
nn
n m
ǫǫ
ǫ