Answer
37
Work Step by Step
By Fermat's little theorem, we can obtain that
$23^{40} \equiv 1 (mod 41)$
Hence,
$23^{1002} = (23^{40})^{25} * 23^{2} \equiv 1 * 529 (mod 41) \equiv 37 (mod 41).$
Discrete Mathematics and Its Applications, Seventh Edition
by Rosen, Kenneth
Published by
McGraw-Hill Education
ISBN 10:
0073383090
ISBN 13:
978-0-07338-309-5
Chapter 4 - Section 4.4 - Solving Congruences - Exercises - Page 285: 34
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