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

Revision as of 18:52, 1 December 2019

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

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 $222 \times 4+37 \times1=\boxed{925}$ .

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 $4\cdot225+1\cdot15+0=\boxed{925}$ .

@@ Line 12: / Line 12: @@
 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 \times 4+37 \times1=\boxed{925}</math>.
--gorefeebuddie
 ==Solution 2==

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

Revision as of 18:52, 1 December 2019

Contents

Problem

Solution

Solution 2

See Also