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

Latest revision as of 00:44, 20 November 2024

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 $a=4, b=10, c=1$ . 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 $4\cdot225+1\cdot15+10=\boxed{925}$ . ~SHEN KISLAY KAI 2023

Solution 3 (Official MAA)

The problem is equivalent to finding a solution to the system of Diophantine equations $196a+14b+c=225a+15c+b$ and $225a+15c+b=216a+36c+6a+c,$ where $1\le a\le 5,\,0\le b\le 13,$ and $0\le c\le 5.$ Simplifying the second equation gives $b=22c-3a.$ Substituting for $b$ in the first equation and simplifying then gives $a=4c,$ so $a = 4$ and $c = 1,$ and the base- $10$ representation of $n$ is $222 \cdot 4 + 37 \cdot 1 = 925.$ It may be verified that $b=10\le 13.$

Solution 4 (Simple Modular Arithmetic)

We're given that $196a+14b+c=225a+15c+b=222a+37c.$ By taking the difference of the first $2$ equalities, we receive $29a+14c=13b.$ Taking $\pmod{13}$ , we receive $3a+c \equiv 0 \pmod{13}.$ We receive the following cases: $(a,c)=(4,1)$ or $(3,4).$ (Note that $(2,7)$ doesn't work since $a,c<6$ by third condition). We can just check these two, and find that $(a,b,c)=(4,10,1),$ and just plugging in $(a,c)$ into the third expression we receive $888+37=\boxed{925}.$

~SirAppel

Video Solution

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

@@ Line 1: / Line 1: @@
+==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==
-==Solutions==
-==Solution Algebra==
 We have these equations:
@@ Line 10: / Line 9: @@
 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*4+37*1=\boxed{925}</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>.
+~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==
+https://www.youtube.com/watch?v=WVtbD8x9fCM
+~Shreyas S
--expiLnCalc
+==See Also==
+{{AIME box|year=2018|n=I|num-b=1|num-a=3}}
+{{MAA Notice}}

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

Page

Toolbox

Search