Difference between revisions of "2018 AIME I Problems/Problem 2"
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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>. | 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>. | ||
| − | - | + | -gorefeebuddie |
==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=1|num-a=3}} | {{AIME box|year=2018|n=I|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:35, 15 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 
.
-gorefeebuddie
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. 
