Difference between revisions of "2012 AIME I Problems/Problem 5"

m (Video Solution by Richard Rusczyk)
Line 5: Line 5:
  
 
When <math>1</math> is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in <math>10.</math> Therefore, every subtraction involving two numbers from <math>B</math> will necessarily involve exactly one number ending in <math>10.</math> To solve the problem, then, we can simply count the instances of such numbers. With the <math>10</math> in place, the seven remaining <math>1</math>'s can be distributed in any of the remaining <math>11</math> spaces, so the answer is <math>{11 \choose 7} = \boxed{330}</math>.
 
When <math>1</math> is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in <math>10.</math> Therefore, every subtraction involving two numbers from <math>B</math> will necessarily involve exactly one number ending in <math>10.</math> To solve the problem, then, we can simply count the instances of such numbers. With the <math>10</math> in place, the seven remaining <math>1</math>'s can be distributed in any of the remaining <math>11</math> spaces, so the answer is <math>{11 \choose 7} = \boxed{330}</math>.
 
=== Video Solution by Richard Rusczyk ===
 
 
https://artofproblemsolving.com/videos/amc/2012aimei/330
 
 
~ dolphin7
 
  
 
==Video Solution==
 
==Video Solution==

Revision as of 18:08, 7 July 2020

Problem 5

Let $B$ be the set of all binary integers that can be written using exactly $5$ zeros and $8$ ones where leading zeros are allowed. If all possible subtractions are performed in which one element of $B$ is subtracted from another, find the number of times the answer $1$ is obtained.

Solution

When $1$ is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in $10.$ Therefore, every subtraction involving two numbers from $B$ will necessarily involve exactly one number ending in $10.$ To solve the problem, then, we can simply count the instances of such numbers. With the $10$ in place, the seven remaining $1$'s can be distributed in any of the remaining $11$ spaces, so the answer is ${11 \choose 7} = \boxed{330}$.

Video Solution

https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s

See also

2012 AIME I (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS