Difference between revisions of "2018 AIME I Problems/Problem 2"
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==Problem== | ==Problem== | ||
− | + | 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>. | |
+ | ==Solution 1== | ||
− | == | + | We have these equations: |
+ | <math>196a+14b+c=225a+15c+b=222a+37c</math>. | ||
+ | Taking the last two we get <math>3a+b=22c</math>. Because <math>c \neq 0</math> otherwise <math>a \ngtr 0</math>, and <math>a \leq 5</math>, <math>c=1</math>. | ||
− | + | Then we know <math>3a+b=22</math>. | |
+ | Taking the first two equations we see that <math>29a+14c=13b</math>. Combining the two gives <math>a=4, b=10, c=1</math>. Then we see that <math>222 \times 4+37 \times1=\boxed{925}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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+10=\boxed{925}</math>. | ||
+ | |||
+ | ==Solution 3 (Official MAA)== | ||
+ | The problem is equivalent to finding a solution to the system of Diophantine equations <math>196a+14b+c=225a+15c+b</math> and <math>225a+15c+b=216a+36c+6a+c,</math> where <math>1\le a\le 5,\,0\le b\le 13,</math> and <math>0\le c\le 5.</math> Simplifying the second equation gives <math>b=22c-3a.</math> Substituting for <math>b</math> in the first equation and simplifying then gives <math>a=4c,</math> so <math>a = 4</math> and <math>c = 1,</math> and the base-<math>10</math> representation of <math>n</math> is <math>222 \cdot 4 + 37 \cdot 1 = 925.</math> It may be verified that <math>b=10\le 13.</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=WVtbD8x9fCM | ||
+ | ~Shreyas S | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2018|n=I|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:38, 26 February 2021
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 1
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 .
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 .
Solution 3 (Official MAA)
The problem is equivalent to finding a solution to the system of Diophantine equations and where and Simplifying the second equation gives Substituting for in the first equation and simplifying then gives so and and the base- representation of is It may be verified that
Video Solution
https://www.youtube.com/watch?v=WVtbD8x9fCM ~Shreyas S
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.