Difference between revisions of "2012 AIME I Problems/Problem 5"
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| − | == Problem | + | == 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 == | ||
| − | When a number | + | 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 Solution== | ||
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| + | https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s | ||
== 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}} | ||
Revision as of 20:18, 24 January 2021
Contents
Problem
Let 
be the set of all binary integers that can be written using exactly 
zeros and 
ones where leading zeros are allowed. If all possible subtractions are performed in which one element of is subtracted from another, find the number of times the answer 
is obtained.
Solution
When is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in 
Therefore, every subtraction involving two numbers from will necessarily involve exactly one number ending in To solve the problem, then, we can simply count the instances of such numbers. With the 
in place, the seven remaining 's can be distributed in any of the remaining 
spaces, so the answer is 
.
Video Solution
https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
