Difference between revisions of "2018 AIME I Problems/Problem 2"
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+ | ==Solution 2== | ||
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+ | We know that <math>196a+14b+c=225a+15c+b=222a+37c</math>. Combining the first and third equations give that <math>196a+14b+c=222a+37c</math>, or <cmath>7b=13a+18c</cmath> | ||
+ | The second and third gives <math>222a+37c=225a+15c+b</math>, or <cmath>22c-3a=b </cmath><cmath> 154c-21a=7b=13a+18c </cmath><cmath> 4c=a</cmath> | ||
+ | We can have <math>a=4,8,12</math>, but only <math>a=4</math> falls within the possible digits of base <math>6</math>. Thus <math>a=4</math>, <math>c=1</math>, and thus you can find <math>b</math> which equals <math>10</math>. Thus, our answer is <math>4\cdot225+1\cdot15+0=\boxed{925}</math>. | ||
==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 23:49, 5 May 2019
Contents
Problem
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 .
Solution
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
Solution 2
We know that . Combining the first and third equations give that , or The second and third gives , or We can have , but only falls within the possible digits of base . Thus , , and thus you can find which equals . Thus, our answer is .
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.