Question 164864: What is the ones's digit of 8^1007?
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Found 2 solutions by Alan3354, gonzo:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! What is the ones's digit of 8^1007?
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The one's digit of any even number (except zero) to a power of a multiple of 4, is 6. That is, 8^(4*n) --> xxxxxxxx6 (n is an integer).
1007 = 4*251 + 3.
8^1007 = (8^1004)*(8^3)
8^3 = 512.
Multiplying an even number by 6 does not change the one's digit.
So, the one's digit of 8^1007 is 2.
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PS All calculators have a limit on their resolution, and they can only generate results to a certain number of significant figures. This does not mean that the digits to the right are actually zeroes, but only that the calculator has a limited ability to generate and display a limited number of digits, and it fills in the space with zeroes.
Answer by gonzo(654)  (Show Source):
You can put this solution on YOUR website! an interesting thing happens to the number as the exponent gets past 16.
the ones digit becomes 0 and stays 0 thereafter.
as you go up in exponent power, the number of 0's on the right hand side of the number keeps expanding.
for example:
8^17 has one zero on the right (unit digit).
8^18 has 3 zeroes on the right (unit, ten, hundred digit).
8^22 has 5 zeroes on the right (unit, ten, hundred, thousand, ten thousand digit)
so, even though my calculator and my excel spreadsheet don't go up that far, i would assume the ones digit will remain 0 after 8^17.
for a while (up to 8^11), the ones digit was repeating (8,4,2,6,8,4,2,6,8,4,2,6) and i was tempted to give you that answer but my curiosity got the best of me after my calculator threw up on 8^12 and i went to the excel spreadsheet. it was there that i saw what was really happening.
here's a display of what the excel spreadsheet was telling me.
the first column is the number raised to the power in the second column.
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x : 8 raised to the power of x
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1 : 8
2 : 64
3 : 512
4 : 4096
5 : 32768
6 : 262144
7 : 2097152
8 : 16777216
9 : 134217728
10 : 1073741824
11 : 8589934592
12 : 68719476736
13 : 549755813888
14 : 4398046511104
15 : 35184372088832
16 : 281474976710656
17 : 2251799813685250
18 : 18014398509482000
19 : 144115188075856000
20 : 1152921504606850000
21 : 9223372036854780000
22 : 73786976294838200000
23 : 590295810358706000000
24 : 4722366482869650000000
25 : 37778931862957200000000
26 : 302231454903657000000000
27 : 2417851639229260000000000
28 : 19342813113834100000000000
29 : 154742504910673000000000000
30 : 1237940039285380000000000000
31 : 9903520314283040000000000000
32 : 79228162514264300000000000000
33 : 633825300114115000000000000000
34 : 5070602400912920000000000000000
35 : 40564819207303300000000000000000
36 : 324518553658427000000000000000000
37 : 2596148429267410000000000000000000
38 : 20769187434139300000000000000000000
39 : 166153499473114000000000000000000000
40 : 1329227995784920000000000000000000000
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i'm not sure what the algebraic answer to this would be.
i would supposes that after a certain amount of raising to a power for any number that the ones digit will eventually start being 0, but i don't know how to explain that other than it happens. for example: 2 raised to the 50th power starts exhibiting the same thing, i.e. the ones digit becomes 0 and remains 0 thereafter.
a similarity in both these instances is that the number of digits other than 0 maxed at around 15.
did the same thing with 3 to the power of whatever and max number of digits before started getting 0's on the right appeared to be 15 as well.
i'm sure there's an explanation.
i just don't know it.
anyway, 0 is your answer to the best of my knowledge.
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