Difference between revisions of "2017 AIME I Problems/Problem 5"
m (→Solution 1) |
|||
| Line 22: | Line 22: | ||
==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> | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://youtu.be/Mk-MCeVjSGc | ||
| + | ~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}} | ||
Revision as of 18:03, 17 June 2020
Problem 5
A rational number written in base eight is 
, where all digits are nonzero. The same number in base twelve is 
. Find the base-ten number 
.
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.
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 
or 
.
Evaluating the places to the right side of the decimal point gives us 
or 
.
When the numbers are converted into base 8, we get 
and 
. Since 
, the first value is correct. Compiling the necessary digits leaves us a final answer of 
Solution 2
The parts before and after the decimal points must be equal. Therefore 
and 
. Simplifying the first equation gives 
. Plugging this into the second equation gives 
. Multiplying both sides by 64 gives 
. 
and 
are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using 
, 
or 
. Testing these gives that doesn't work, and 
gives 
, and 
. Therefore 
Video Solution
https://youtu.be/Mk-MCeVjSGc ~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. 
