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

Revision as of 21:10, 2 May 2023

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 tens 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 $a = (3/2)b$ . 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 $0$ , therefor the answer is $\boxed{321}$

By maxamc

Video Solution

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

@@ Line 22: / Line 22: @@
 ==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>
+==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>, therefor the answer is <math>\boxed{321}</math>
+By maxamc
 ==Video Solution==

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

Page

Toolbox

Search