Difference between revisions of "2012 AIME I Problems/Problem 5"
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== 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>. |
== 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}} |
Revision as of 02:01, 17 March 2012
Problem 5
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
.
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 |