For the last three digits:
7^20 = 79792266297612001,
which is congruent to 1 modulo 1000. This means that the last three digits cycle every 20 times.
So 7^2005 = 7^(2000 + 5) = (7^20)^100 * 7^5,
so both 7^2005 and 7^5 have the same last three digits.
Thus the last three digits of 7^2005 are 807, since 7^5 = 16807.
2007-11-05 07:07:43
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answer #1
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answered by ♣ K-Dub ♣ 6
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You should have given us the complete question. If it asks what are the last three digits of 7^2005, it was wrong of you to suppose that the first step was to calculate 7^2005, and the second step to throw away all but the last three digits, but I think that is exactly what you supposed.
You have been shown by another answerer that the last 3 digits of 7^20 are ...001, and how the simple route to the answer follows from that. Whether the question is 7^2005, 7^865, 7^45 or anything else, the answer is the same.
There is a theorem of Euler's from which you can deduce that the last 3 digits of 7^100 are ...001, after which it is trial and error to see how many factors of 2 or 5 you can discard for that still to be true, discovering by three trials and two errors that you can discard one 5 but no 2's.
2007-11-05 15:49:34
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answer #2
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answered by bh8153 7
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The value is approximately 2.639795 * 10^1694
I obtained this by calculating 2005*log_10(7) (that is, the base-10 log of 7^2005). This is approximately 1694.42157022
Thus, 7^2005 â 10^1694.42157022 = (10^1694)(10^0.42157022) â 2.639795 * 10^1694
The exact value would have 1695 digits. A symbolic algebra program like Mathematica can calculate the exact value. So can bc, a built-in program available on most Unix/Linux computers.
2007-11-05 12:55:11
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answer #3
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answered by Ron W 7
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There is no way to give an exact value for this expression that is faster than just multiplying 7 by itself, 2005 times. The result will probably overflow your calculator; you will have to use infinite-precision math software such as Mathematica to get an exact numerical answer.
It is possible to get an approximate answer, however, by using logarithms:
7^2005 = 10^(2005*log10(7),
so compute 2005*log10(7), and that is the log base 10 of the answer. In other words, 10 to that exponent is the approximate answer.
2007-11-05 12:45:19
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answer #4
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answered by acafrao341 5
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I believe they are asking the first digit, i remember these kind of problems.
7^1 = 7
7^2 = 49 first digit is 9
7^3 = 49*7 = smth that ends with 3 (9*7=63)
7^4 = ..3*7 = smth that ends with 1 (3*7=21)
7^5 = ..7
7^6 = ..9
7^7 = ..3
7^8 = ..1
See the pattern ? This repeats on and on.
7^4 = ..1
7^8 = ..1
7^(2004) =..1
7^2007 = ..7
2007-11-05 12:43:36
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answer #5
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answered by zadig 2
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