Each of the individual power terms has a cycle length of 1, 2 or 4, so the overall cycle length must be one of these, and is in fact 4 as shown below:
i^n mod 10
n i = 1 2 3 4 5 6 7 8 9 A mod 10
1 1 2 3 4 5 6 7 8 9 5
2 1 4 9 6 5 6 9 4 1 5
3 1 8 7 4 5 6 3 2 9 5
4 1 6 1 6 5 6 1 6 1 3
5 1 2 3 4 5 6 7 8 9 5
6 1 4 9 6 5 6 9 4 1 5
7 1 8 7 4 5 6 3 2 9 5
8 1 6 1 6 5 6 1 6 1 3
9 1 2 3 4 5 6 7 8 9 5
10 1 4 9 6 5 6 9 4 1 5
11 1 8 7 4 5 6 3 2 9 5
12 1 6 1 6 5 6 1 6 1 3
13 1 2 3 4 5 6 7 8 9 5
14 1 4 9 6 5 6 9 4 1 5
15 1 8 7 4 5 6 3 2 9 5
16 1 6 1 6 5 6 1 6 1 3
17 1 2 3 4 5 6 7 8 9 5
18 1 4 9 6 5 6 9 4 1 5
19 1 8 7 4 5 6 3 2 9 5
20 1 6 1 6 5 6 1 6 1 3
21 1 2 3 4 5 6 7 8 9 5
22 1 4 9 6 5 6 9 4 1 5
23 1 8 7 4 5 6 3 2 9 5
24 1 6 1 6 5 6 1 6 1 3
25 1 2 3 4 5 6 7 8 9 5
The above table was produced with
v1 = 1
v2 = 2
v3 = 3
v4 = 4
v5 = 5
v6 = 6
v7 = 7
v8 = 8
v9 = 9
CLS
FOR n = 1 TO 25
PRINT n,
PRINT v1 MOD 10;
PRINT v2 MOD 10;
PRINT v3 MOD 10;
PRINT v4 MOD 10;
PRINT v5 MOD 10;
PRINT v6 MOD 10;
PRINT v7 MOD 10;
PRINT v8 MOD 10;
PRINT v9 MOD 10,
PRINT (v1 + v2 + v3 + v4 + v5 + v6 + v7 + v8 + v9) MOD 10
v1 = v1 * 1 MOD 10
v2 = v2 * 2 MOD 10
v3 = v3 * 3 MOD 10
v4 = v4 * 4 MOD 10
v5 = v5 * 5 MOD 10
v6 = v6 * 6 MOD 10
v7 = v7 * 7 MOD 10
v8 = v8 * 8 MOD 10
v9 = v9 * 9 MOD 10
NEXT
So, when n is a multiple of 4, A = 3, otherwise A = 5.
That can be turned into an explicit formula:
A = 3 - 2*[[n/4]-n/4]
where [x] = largest integer smaller than x, i.e., the FLOOR function, so that, for example, [-.25] = -1.
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Posted by Charlie
on 2011-02-23 20:57:50 |
solution
