Difference between revisions of "2018 AIME I Problems/Problem 2"
(→Solution 2) |
|||
(10 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
− | |||
− | |||
==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>. | 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== | + | ==Solution 1== |
We have these equations: | We have these equations: | ||
Line 11: | Line 9: | ||
Then we know <math>3a+b=22</math>. | Then we know <math>3a+b=22</math>. | ||
− | Taking the first two equations we see that <math>29a+ | + | 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== | ==Solution 2== | ||
Line 17: | Line 15: | ||
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> | 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> | 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> | + | 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>. |
+ | ~SHEN KISLAY KAI 2023 | ||
+ | |||
+ | ==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> | ||
+ | ==Solution 4 (Simple Modular Arithmetic)== | ||
+ | We're given that <math>196a+14b+c=225a+15c+b=222a+37c.</math> By taking the difference of the first <math>2</math> equalities, we receive <math>29a+14c=13b.</math> Taking <math>\pmod{13}</math>, we receive <math>3a+c \equiv 0 \pmod{13}.</math> We receive the following cases: <math>(a,c)=(4,1)</math> or <math>(3,4).</math> (Note that <math>(2,7)</math> doesn't work since <math>a,c<6</math> by third condition). We can just check these two, and find that <math>(a,b,c)=(4,10,1),</math> and just plugging in <math>(a,c)</math> into the third expression we receive <math>888+37=\boxed{925}.</math> | ||
+ | |||
+ | ~SirAppel | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 00:44, 20 November 2024
Contents
[hide]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 . ~SHEN KISLAY KAI 2023
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
Solution 4 (Simple Modular Arithmetic)
We're given that By taking the difference of the first equalities, we receive Taking , we receive We receive the following cases: or (Note that doesn't work since by third condition). We can just check these two, and find that and just plugging in into the third expression we receive
~SirAppel
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.