Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 4"

(New page: == Problem == When <math>1 + 7 + 7^2 + \cdots + 7^{2004}</math> is divided by <math>1000</math>, a remainder of <math>N</math> is obtained. Determine the value of <math>N</math>. == Solut...)
(No difference)

Revision as of 16:36, 21 March 2008

Problem

When $1 + 7 + 7^2 + \cdots + 7^{2004}$ is divided by $1000$, a remainder of $N$ is obtained. Determine the value of $N$.

Solution

By the geometric series formula, $1 + 7 + 7^2 + \cdots + 7^{2004} = \frac{7^{2005}-1}{7-1} = \frac{7^{2005}-1}{6}$. Since $\varphi(1000) = 400$, by Fermat-Euler's Theorem, this is equivalent to finding $\frac{7^{400 \cdot 5 + 6} - 1}{6} \equiv \frac{7^5 - 1}{6} \equiv \boxed{810} \pmod{1000}$.

See also

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by
Problem 3
Followed by
Problem 5
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