Difference between revisions of "2018 AIME I Problems/Problem 2"
Elephant353 (talk | contribs) m (→Solution Algebra: Setting up a navigational box for the 2018 AIME I problems.) |
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Then we know <math>3a+b=22</math>. | Then we know <math>3a+b=22</math>. | ||
− | Taking the first two equations we see that <math>29a+14=13b</math>. Combining the two gives <math>a=4, b=10</math>. Then we see that <math>222 | + | Taking the first two equations we see that <math>29a+14=13b</math>. Combining the two gives <math>a=4, b=10</math>. Then we see that <math>222 \times 4+37 \times1=\boxed{925}</math>. |
-expiLnCalc | -expiLnCalc |
Revision as of 21:40, 9 March 2018
The number can be written in base as , can be written in base as , and can be written in base as , where . Find the base- representation of .
Solutions
Solution Algebra
We have these equations: . Taking the last two we get . Because otherwise , and , .
Then we know . Taking the first two equations we see that . Combining the two gives . Then we see that .
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See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.