Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AIME I Problems/Problem 5"

(→‎Solution 1)
m (→‎Solution 1)
 
(8 intermediate revisions by 6 users not shown)
Line 3: Line 3:
  
 
==Solution 1==
 
==Solution 1==
−
First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base twelve that have equal twelves and ones digits in base 8.
+
First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12.
  
 
<math>11_{12}=15_8</math>
 
<math>11_{12}=15_8</math>
Line 18: Line 18:
 
Compare the ones places to check if they are equal. We find that they are equal if <math>b=2</math> or <math>b=4</math>.
 
Compare the ones places to check if they are equal. We find that they are equal if <math>b=2</math> or <math>b=4</math>.
 
Evaluating the places to the right side of the decimal point gives us <math>22.23_{12}</math> or <math>44.46_{12}</math>.
 
Evaluating the places to the right side of the decimal point gives us <math>22.23_{12}</math> or <math>44.46_{12}</math>.
−
When the numbers are converted into base 8, we get <math>32.14_8</math> and <math>64.30_8</math>. Since <math>d\neq0</math>, the first value is correct. Compiling the necessary digits leaves us a final answer of <math>\boxed{321}</math>  
+
When the numbers are converted into base 8, we get <math>32.14_8</math> and <math>64.30_8</math>. Since <math>d\neq0</math>, the first value is correct. Compiling the necessary digits leaves us a final answer of <math>\boxed{321}</math>
−
 
 
−
<math>\blacksquare</math><br>
 
−
[[User:a1b2|a1b2]]
 
  
 
==Solution 2==
 
==Solution 2==
 
The parts before and after the decimal points must be equal. Therefore <math>8a + b = 12b + b</math> and <math>c/8 + d/64 = b/12 + a/144</math>.  Simplifying the first equation gives <math>a = (3/2)b</math>.  Plugging this into the second equation gives <math>3b/32 = c/8 + d/64</math>. Multiplying both sides by 64 gives <math>6b = 8c + d</math>.  <math>a</math> and <math>b</math> are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using <math>a = 3/2b</math>, <math>(a,b) = (3,2)</math> or <math>(6,4)</math>.  Testing these gives that <math>(6,4)</math> doesn't work, and <math>(3,2)</math> gives <math>a = 3, b = 2, c = 1</math>, and <math>d = 4</math>.  Therefore <math>abc = \boxed{321}</math>
 
The parts before and after the decimal points must be equal. Therefore <math>8a + b = 12b + b</math> and <math>c/8 + d/64 = b/12 + a/144</math>.  Simplifying the first equation gives <math>a = (3/2)b</math>.  Plugging this into the second equation gives <math>3b/32 = c/8 + d/64</math>. Multiplying both sides by 64 gives <math>6b = 8c + d</math>.  <math>a</math> and <math>b</math> are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using <math>a = 3/2b</math>, <math>(a,b) = (3,2)</math> or <math>(6,4)</math>.  Testing these gives that <math>(6,4)</math> doesn't work, and <math>(3,2)</math> gives <math>a = 3, b = 2, c = 1</math>, and <math>d = 4</math>.  Therefore <math>abc = \boxed{321}</math>
 +
 +
==Solution 3==
 +
Converting to base <math>10</math> we get
 +
 +
<math>4604a+72c+9d=6960b</math>
 +
 +
Since <math>72c</math> and <math>9d</math> are much smaller than the other two terms, dividing by <math>100</math> and approximating we get
 +
 +
<math>46a=70b</math>
 +
 +
Writing out the first few values of <math>a</math> and <math>b</math>, the first possible tuple is
 +
 +
<math>a=3, b=2, c=1, d=4</math>
 +
 +
and the second possible tuple is
 +
 +
<math>a=6, b=4, c=3, d=0</math>
 +
 +
Note that <math>d</math> can not be <math>0</math>, therefore the answer is <math>\boxed{321}</math>
 +
 +
By maxamc
 +
 +
==Solution 4 (modular congruency/nt bash)==
 +
In the problem, we are given that <cmath>8a+b+\dfrac c8+\dfrac d{64}=12b+b+\dfrac b{12}+\dfrac a{144}.</cmath> We multiply by the LCM of the denominators, which is <math>576</math> to get <cmath>4608a+576b+72c+9d=6912b+576b+48b+4a.</cmath> We then group like terms and factor to get <cmath>4604a+72c+9d=48b(144+1)=145\cdot48b=24\cdot290b.</cmath>
 +
 +
Observe that the coefficients of <math>a</math>, <math>c</math>, and <math>b</math> are all divisible by <math>4</math>. Therefore, we know that <math>d</math> must also be divisible by <math>4</math> to compensate. (To observe this, one could rearrange the terms to see <math>9d=24\cdot290b-4604a-72c=4(6\cdot290b-1151a-18c)</math>. At this point, it is obvious that <math>9d</math> must be divisible by <math>4</math>, so <math>d</math> must be divisible by <math>4</math>.) Thus, we let <math>d=4y</math> to see <math>4604a+72c+9\cdot4y=24\cdot290b</math>. We divide by <math>4</math> in the equation to get <cmath>1151a+18c+9y=6\cdot290b.</cmath>
 +
 +
Observe that the coefficients of <math>c</math>, <math>y</math>, and <math>b</math> are all divisible by <math>3</math>. By similar reasoning, <math>a</math> must be divisible by <math>3</math>, so we let <math>a=3x</math>. Substituting and dividing by <math>3</math>, we get <cmath>1151x+6c+3y=2\cdot290b.</cmath> We observe that this can no longer be reduced by similar means.
 +
 +
We know that <math>0<a,b,c<8\implies0<3x,b,4y<8</math>. We examine <math>0<4y<8</math>; this becomes <math>0<y<2</math>. It is apparent that <math>y=1\implies d=4</math>. However, the problem does not even ask for <math>d</math>, so it may appear that this find is meaningless. However, we substitute our value of <math>y</math> in to get <cmath>1151x+6c+3=580b.</cmath>
 +
 +
We know that <math>x</math> is either <math>1</math> or <math>2</math>, since <math>0<3x<8</math>. We take our equation modulo <math>6</math> to find <cmath>5x+3\equiv4b\pmod{6}\implies3\equiv4b+x\pmod{6}.</cmath> If <math>x=2</math>, then <math>1\equiv4b\pmod{6}</math> - this is equivalent to saying that <math>1+6n=4b\implies 1=4b-6n</math>. However, <math>4b-6n</math> is always even, and <math>1</math> is odd. Thus, this case is not possible, so <math>x=1\implies a=3</math>. We know that <math>x=1</math>; thus, <math>3\equiv4b+1\pmod{6}\implies2\equiv4b\pmod{6}\implies1\equiv2b\pmod{3}</math>. Obviously, <math>b=2</math> and <math>b=5</math> work; however, if <math>b=5</math>, then the RHS of our original equation (<math>2\cdot290b</math>) is much too large to be equal to the LHS (the maximum possible value of the LHS is <math>1151\cdot1+6\cdot7+3\cdot1</math>, which is less than <math>1200</math> while the RHS would become <math>2\cdot290\cdot5=2900</math>), so we have <math>b=2</math>.
 +
 +
We recall our equation of <math>1151x+6c+3=580b</math>. Plugging in what we know, we have <math>1151\cdot1+6c+3=580\cdot2\implies6c=6\implies c=1</math>.
 +
 +
Therefore, <math>\overline{abc}=\boxed{321}</math>.
 +
 +
~~this insane timesink solution was composed by Technodoggo~~
 +
 +
==Video Solution==
 +
 +
https://youtu.be/Mk-MCeVjSGc?t=298
 +
~Shreyas S
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2017|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2017|n=I|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:03, 18 January 2024

Problem 5

A rational number written in base eight is $\underline{ab} . \underline{cd}$, where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$. Find the base-ten number $\underline{abc}$.

Solution 1

First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12.

$11_{12}=15_8$

$22_{12}=32_8$

$33_{12}=47_8$

$44_{12}=64_8$

$55_{12}=101_8$

We stop because we only can have two-digit numbers in base 8 and 101 is not a 2 digit number. Compare the ones places to check if they are equal. We find that they are equal if $b=2$ or $b=4$. Evaluating the places to the right side of the decimal point gives us $22.23_{12}$ or $44.46_{12}$. When the numbers are converted into base 8, we get $32.14_8$ and $64.30_8$. Since $d\neq0$, the first value is correct. Compiling the necessary digits leaves us a final answer of $\boxed{321}$

Solution 2

The parts before and after the decimal points must be equal. Therefore $8a + b = 12b + b$ and $c/8 + d/64 = b/12 + a/144$. Simplifying the first equation gives $a = (3/2)b$. Plugging this into the second equation gives $3b/32 = c/8 + d/64$. Multiplying both sides by 64 gives $6b = 8c + d$. $a$ and $b$ are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using $a = 3/2b$, $(a,b) = (3,2)$ or $(6,4)$. Testing these gives that $(6,4)$ doesn't work, and $(3,2)$ gives $a = 3, b = 2, c = 1$, and $d = 4$. Therefore $abc = \boxed{321}$

Solution 3

Converting to base $10$ we get

$4604a+72c+9d=6960b$

Since $72c$ and $9d$ are much smaller than the other two terms, dividing by $100$ and approximating we get

$46a=70b$

Writing out the first few values of $a$ and $b$, the first possible tuple is

$a=3, b=2, c=1, d=4$

and the second possible tuple is

$a=6, b=4, c=3, d=0$

Note that $d$ can not be $0$, therefore the answer is $\boxed{321}$

By maxamc

Solution 4 (modular congruency/nt bash)

In the problem, we are given that \[8a+b+\dfrac c8+\dfrac d{64}=12b+b+\dfrac b{12}+\dfrac a{144}.\] We multiply by the LCM of the denominators, which is $576$ to get \[4608a+576b+72c+9d=6912b+576b+48b+4a.\] We then group like terms and factor to get \[4604a+72c+9d=48b(144+1)=145\cdot48b=24\cdot290b.\]

Observe that the coefficients of $a$, $c$, and $b$ are all divisible by $4$. Therefore, we know that $d$ must also be divisible by $4$ to compensate. (To observe this, one could rearrange the terms to see $9d=24\cdot290b-4604a-72c=4(6\cdot290b-1151a-18c)$. At this point, it is obvious that $9d$ must be divisible by $4$, so $d$ must be divisible by $4$.) Thus, we let $d=4y$ to see $4604a+72c+9\cdot4y=24\cdot290b$. We divide by $4$ in the equation to get \[1151a+18c+9y=6\cdot290b.\]

Observe that the coefficients of $c$, $y$, and $b$ are all divisible by $3$. By similar reasoning, $a$ must be divisible by $3$, so we let $a=3x$. Substituting and dividing by $3$, we get \[1151x+6c+3y=2\cdot290b.\] We observe that this can no longer be reduced by similar means.

We know that $0<a,b,c<8\implies0<3x,b,4y<8$. We examine $0<4y<8$; this becomes $0<y<2$. It is apparent that $y=1\implies d=4$. However, the problem does not even ask for $d$, so it may appear that this find is meaningless. However, we substitute our value of $y$ in to get \[1151x+6c+3=580b.\]

We know that $x$ is either $1$ or $2$, since $0<3x<8$. We take our equation modulo $6$ to find \[5x+3\equiv4b\pmod{6}\implies3\equiv4b+x\pmod{6}.\] If $x=2$, then $1\equiv4b\pmod{6}$ - this is equivalent to saying that $1+6n=4b\implies 1=4b-6n$. However, $4b-6n$ is always even, and $1$ is odd. Thus, this case is not possible, so $x=1\implies a=3$. We know that $x=1$; thus, $3\equiv4b+1\pmod{6}\implies2\equiv4b\pmod{6}\implies1\equiv2b\pmod{3}$. Obviously, $b=2$ and $b=5$ work; however, if $b=5$, then the RHS of our original equation ($2\cdot290b$) is much too large to be equal to the LHS (the maximum possible value of the LHS is $1151\cdot1+6\cdot7+3\cdot1$, which is less than $1200$ while the RHS would become $2\cdot290\cdot5=2900$), so we have $b=2$.

We recall our equation of $1151x+6c+3=580b$. Plugging in what we know, we have $1151\cdot1+6c+3=580\cdot2\implies6c=6\implies c=1$.

Therefore, $\overline{abc}=\boxed{321}$.

~~this insane timesink solution was composed by Technodoggo~~

Video Solution

https://youtu.be/Mk-MCeVjSGc?t=298 ~Shreyas S

See Also

2017 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_5&oldid=211210"