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

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== Problem 5 ==
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== Problem ==
 
Let <math>B</math> be the set of all binary integers that can be written using exactly <math>5</math> zeros and <math>8</math> ones where leading zeros are allowed. If all possible subtractions are performed in which one element of <math>B</math> is subtracted from another, find the number of times the answer <math>1</math> is obtained.
 
Let <math>B</math> be the set of all binary integers that can be written using exactly <math>5</math> zeros and <math>8</math> ones where leading zeros are allowed. If all possible subtractions are performed in which one element of <math>B</math> is subtracted from another, find the number of times the answer <math>1</math> is obtained.
  
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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 ===
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==Video Solutions==
  
https://artofproblemsolving.com/videos/amc/2012aimei/330
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https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s
 
 
~ dolphin7
 
 
 
==Video Solution==
 
  
https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s
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https://www.youtube.com/watch?v=f5ZoAFfc-1E&list=PLyhPcpM8aMvIo_foUDwmXnQClMHngjGto&index=5 (Solution by Richard Rusczyk) - AMBRIGGS
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2012|n=I|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:14, 30 July 2022

Problem

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 Solutions

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

https://www.youtube.com/watch?v=f5ZoAFfc-1E&list=PLyhPcpM8aMvIo_foUDwmXnQClMHngjGto&index=5 (Solution by Richard Rusczyk) - AMBRIGGS

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

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