Mock AIME 1 Pre 2005 Problems/Problem 4: Difference between revisions
Pianoforte (talk | contribs) final answer corrected |
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== Solution == | == Solution == | ||
By the [[geometric series]] formula, <math>1 + 7 + 7^2 + \cdots + 7^{2004} = \frac{7^{2005}-1}{7-1} = \frac{7^{2005}-1}{6}</math>. Since <math>\varphi(1000) = 400</math>, by [[Fermat's Little Theorem|Fermat-Euler's Theorem]], this is equivalent to finding <math>\frac{7^{400 \cdot 5 + | By the [[geometric series]] formula, <math>1 + 7 + 7^2 + \cdots + 7^{2004} = \frac{7^{2005}-1}{7-1} = \frac{7^{2005}-1}{6}</math>. Since <math>\varphi(1000) = 400</math>, by [[Fermat's Little Theorem|Fermat-Euler's Theorem]], this is equivalent to finding <math>\frac{7^{400 \cdot 5 + 5} - 1}{6} \equiv \frac{7^5 - 1}{6} \equiv \boxed{801} \pmod{1000}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 09:51, 11 June 2013
Problem
When 
is divided by 
, a remainder of 
is obtained. Determine the value of .
Solution
By the geometric series formula, 
. Since 
, by Fermat-Euler's Theorem, this is equivalent to finding 
.
See also
| Mock AIME 1 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
