2012 AIME I Problems/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.


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=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
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All AIME Problems and Solutions

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