Difference between revisions of "2021 JMPSC Sprint Problems/Problem 15"
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== Solution 2 == | == Solution 2 == | ||
− | 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...\boxed{75}</math>. |
+ | ~Mathdreams | ||
+ | |||
+ | == Solution 3 == | ||
+ | <cmath>100^{10} \equiv 0 \mod 100</cmath><cmath>5^{10} \equiv 25 \mod 100</cmath>Therefore, the answer is <math>75</math> | ||
+ | |||
+ | - kante314 - | ||
− | + | ==See also== | |
+ | #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]] | ||
+ | #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 19:54, 7 September 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
Solution 3
Therefore, the answer is
- kante314 -
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.