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"

m (→‎See Also)
(→‎Solution 2)
Line 22: Line 22:
  
 
==Solution 2==
 
==Solution 2==
−
The parts before the decimal points must be equal as must the parts after. Therefore 8<math>a</math> + <math>b</math> = 12<math>b</math> + <math>b</math> and <math>c</math>/8 + <math>d</math>/64 = <math>b</math>/12 + <math>a</math>/144.  Simplifying the first equation gives: <math>a</math> = 3/2<math>b</math>.  Plugging this into the second equation gives 3<math>b</math>/32 = <math>c</math>/8 + <math>d</math>/64. Multiply by 64: 6<math>b</math> = 8<math>c</math> + <math>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</math> = 3/2<math>b</math>, <math>a</math>,<math>b</math> = (3,2) or (6,4).  Testing these gives that (6,4) doesn't work, and (3,2) gives <math>a</math> = 3, <math>b</math> = 2, <math>c</math> = 1, and <math>d</math> = 4.  Therefore <math>abc</math> = <math>\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 = 3/2b</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>
  
 
==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}}

Revision as of 20:00, 9 March 2017

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 twelve that have equal twelves and ones digits in base 8.

$11_{12}=15_8$

$22_{12}=32_8$

$33_{12}=47_8$

$44_{12}=64_8$

$55_{12}=101_8$

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 $a = 3/2b$. 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 $abc = \boxed{321}$

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=84580"