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someone can help me solve this problem? 7^2005

2007-11-05 04:32:46 · 5 answers · asked by h0pefree 1 in Science & Mathematics Mathematics

and find the last 3 digits

2007-11-05 04:56:50 · update #1

5 answers

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 · answer #1 · answered by ♣ K-Dub ♣ 6 · 1 0

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 · answer #2 · answered by bh8153 7 · 0 0

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 ≈ 10^1694.42157022 = (10^1694)(10^0.42157022) ≈ 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 · answer #3 · answered by Ron W 7 · 0 0

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 · answer #4 · answered by acafrao341 5 · 0 0

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 · answer #5 · answered by zadig 2 · 0 0

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