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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.
  
 
== Solution ==
 
== Solution ==
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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>.
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==Video Solutions==
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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}}
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{{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 (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

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