Difference between revisions of "2012 AIME I Problems/Problem 5"
m (→Video Solution by Richard Rusczyk) |
|||
Line 5: | Line 5: | ||
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== | ==Video Solution== |
Revision as of 18:08, 7 July 2020
Contents
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 .
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.