Difference between revisions of "2018 AIME I Problems/Problem 2"

(Solution Algebra)
(Solution Algebra)
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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>.
  
-expiLnCalc
+
-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 22:35, 15 March 2018

The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$, can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$, and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$, where $a > 0$. Find the base-$10$ representation of $n$.

Solutions

Solution Algebra

We have these equations: $196a+14b+c=225a+15c+b=222a+37c$. Taking the last two we get $3a+b=22c$. Because $c \neq 0$ otherwise $a \ngtr 0$, and $a \leq 5$, $c=1$.

Then we know $3a+b=22$. Taking the first two equations we see that $29a+14=13b$. Combining the two gives $a=4, b=10$. Then we see that $222 \times 4+37 \times1=\boxed{925}$.

-gorefeebuddie

See Also

2018 AIME I (ProblemsAnswer KeyResources)
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

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