2012 AIME II Problems/Problem 7
Problem 7
Let be the increasing sequence of positive integers whose binary representation has exactly ones. Let be the 1000th number in . Find the remainder when is divided by .
Solution
Okay, an exercise in counting (lots of binomials to calculate!). In base 2, the first number is , which is the only way to choose 8 1's out of 8 spaces, or . What about 9 spaces? Well, all told, there are , which includes the first 1. Similarly, for 10 spaces, there are which includes the first 9. For 11 spaces, there are , which includes the first 45. You're getting the handle. For 12 spaces, there are , which includes the first 165; for 13 spaces, there are , so we now know that has exactly 13 spaces, so the digit is 1.
Now we just proceed with the other 12 spaces with 7 1's, and we're looking for the number. Well, , so we know that the digit also is 1, and we're left with finding the number with 11 spaces and 6 1's. Now which is too big, but Thus, the digit is 1, and we're now looking for the number with 9 spaces and 5 1's. Continuing the same process, , so the digit is 1, and we're left to look for the number with 8 spaces and 4 1's. But here , so N must be the last or largest 7-digit number with 4 1's. Thus the last 8 digits of must be , and to summarize, in base . Therefore, , and the answer is .
See Also
Video Solution: https://www.youtube.com/watch?v=WRP--6Db1qw
2012 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 8 | |
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