2021 JMPSC Sprint Problems/Problem 15: Difference between revisions
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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>. | ||
~Mathdreams | ~Mathdreams | ||
Revision as of 11:22, 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
