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

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The number <math>n</math> can be written in base <math>14</math> as <math>\underline{a}\text{ }\underline{b}\text{ }\underline{c}</math>, can be written in base <math>15</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{b}</math>, and can be written in base <math>6</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }</math>, where <math>a > 0</math>. Find the base-<math>10</math> representation of <math>n</math>.
 
The number <math>n</math> can be written in base <math>14</math> as <math>\underline{a}\text{ }\underline{b}\text{ }\underline{c}</math>, can be written in base <math>15</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{b}</math>, and can be written in base <math>6</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }</math>, where <math>a > 0</math>. Find the base-<math>10</math> representation of <math>n</math>.
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==Solution==
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==Solution 1==
  
 
We have these equations:
 
We have these equations:
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Then we know <math>3a+b=22</math>.
 
Then we know <math>3a+b=22</math>.
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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>.
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Taking the first two equations we see that <math>29a+14c=13b</math>. Combining the two gives <math>a=4, c=10</math>. Then we see that <math>222 \times 4+37 \times1=\boxed{925}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 19:21, 31 December 2020

Problem

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 $n$.

Solution 1

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

Solution 2

We know that $196a+14b+c=225a+15c+b=222a+37c$. Combining the first and third equations give that $196a+14b+c=222a+37c$, or \[7b=13a+18c\] The second and third gives $222a+37c=225a+15c+b$, or \[22c-3a=b\]\[154c-21a=7b=13a+18c\]\[4c=a\] We can have $a=4,8,12$, but only $a=4$ falls within the possible digits of base $6$. Thus $a=4$, $c=1$, and thus you can find $b$ which equals $10$. Thus, our answer is $4\cdot225+1\cdot15+10=\boxed{925}$.

Video Solution

https://www.youtube.com/watch?v=WVtbD8x9fCM ~Shreyas S

See Also

2018 AIME I (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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