Consider all the powers from
1 to 10 of the numbers one to 10.
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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2
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2
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4
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8
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16
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32
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64
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128
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256
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512
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1024
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3
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3
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9
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27
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81
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243
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729
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2187
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6561
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19683
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59049
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4
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4
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16
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64
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256
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1024
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4096
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16384
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65536
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262144
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0148576
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5
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5
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25
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125
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625
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3125
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15625
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78125
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390625
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1953125
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9765625
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6
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6
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36
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216
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1296
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7776
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46656
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279936
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16796160
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10077696
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60466176
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7
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7
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49
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343
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2401
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16807
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117649
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823543
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5764801
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40353607
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282475249
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8
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8
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64
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512
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4096
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32768
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262144
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2097152
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16777216
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134217728
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1073741824
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9
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9
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81
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729
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6561
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59049
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531441
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4782969
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43046721
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387420489
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3486784401
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10
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10
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100
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1000
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10000
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100000
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1000000
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10000000
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100000000
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1000000000
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10000000000
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We are only interested on
the last digit of each entry in the table above. These are shown in
the table below.
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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2
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2
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4
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8
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6
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2
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4
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8
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6
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2
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4
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3
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3
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9
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7
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1
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3
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9
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7
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1
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3
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9
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4
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4
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6
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4
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6
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4
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6
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4
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6
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4
|
6
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5
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5
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5
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5
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5
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5
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5
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5
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5
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25
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25
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6
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6
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6
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6
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6
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6
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6
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6
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6
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6
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6
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7
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7
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9
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3
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1
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7
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9
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3
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1
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7
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9
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8
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8
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4
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2
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6
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8
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4
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2
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6
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8
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4
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9
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9
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1
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9
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1
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9
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1
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9
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1
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9
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1
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10
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10
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100
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1000
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10000
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100000
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1000000
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10000000
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100000000
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1000000000
|
10000000000
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Notice that the last digits ofandare
constant – they are 1 and 6 respectively for alland
that 6-1=5.
Notice that the last digits ofandmake
cycles of length 4: they are 2,4,8,6 and 7,9,3,1 respectively, and
that 7-2 =5.
Notice that the last digits ofandmake
cycles of length 4: they are 3,9,7,1 and 8,4,2,6 respectively, and
that 8-3 =5.
Notice that the last digits ofandmake
cycles of length 2: they are 4,6 and 9,1 respectively, and that 9-4
=5.
These patterns exist because ifthenso
the last digits are related by a difference of 5 at each stage. If
one oforhas
a cycle of a certain length, so must the other.
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