Difference between revisions of "2012 AIME II Problems/Problem 7"
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== Video Solution == | == Video Solution == | ||
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== See Also == | == See Also == |
Latest revision as of 20:09, 19 July 2024
Contents
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
.
Video Solution
See Also
Video Solution: https://www.youtube.com/watch?v=WRP--6Db1qw
2012 AIME II (Problems • Answer Key • Resources) | ||
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