Difference between revisions of "2021 JMPSC Sprint Problems/Problem 15"
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By multiplying out several powers of <math>5</math>, we can observe that the last <math>2</math> digits are always <math>25</math> (with the exception of <math>5^n</math> where <math>n \le 1</math>). Also, <math>10^10</math> ends with several zeros, so the answer is <math>100...00 - 25 = 99...99 - 24 = 999...75</math>. | By multiplying out several powers of <math>5</math>, we can observe that the last <math>2</math> digits are always <math>25</math> (with the exception of <math>5^n</math> where <math>n \le 1</math>). Also, <math>10^10</math> ends with several zeros, so the answer is <math>100...00 - 25 = 99...99 - 24 = 999...75</math>. | ||
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~Mathdreams | ~Mathdreams |
Revision as of 12:22, 11 July 2021
Problem
Find the last two digits of
Solution
Note that and
.
.
Solution 2
By multiplying out several powers of , we can observe that the last
digits are always
(with the exception of
where
). Also,
ends with several zeros, so the answer is
.
~Mathdreams