Difference between revisions of "2003 AIME I Problems/Problem 13"
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For <math> k=5,\ldots , 10</math>, we end on <math> \binom{10}{k}</math> - we don't want to consider numbers with more than 11 digits. So for each <math> k</math> we get | For <math> k=5,\ldots , 10</math>, we end on <math> \binom{10}{k}</math> - we don't want to consider numbers with more than 11 digits. So for each <math> k</math> we get | ||
− | <math> \binom{k}{k}+\binom{k+1}{k}+\binom{10}{k}=\binom{11}{k+1}</math> | + | <math> \binom{k}{k}+\binom{k+1}{k}+\ldots+\binom{10}{k}=\binom{11}{k+1}</math> |
again by the Hockey Stick Identity. So we get | again by the Hockey Stick Identity. So we get |
Revision as of 09:42, 29 September 2018
Contents
[hide]Problem
Let be the number of positive integers that are less than or equal to and whose base- representation has more 's than 's. Find the remainder when is divided by .
Solution 1
In base- representation, all positive numbers have a leftmost digit of . Thus there are numbers that have digits in base notation, with of the digits being 's.
In order for there to be more 's than 's, we must have . Therefore, the number of such numbers corresponds to the sum of all numbers on or to the right of the vertical line of symmetry in Pascal's Triangle, from rows to (as ). Since the sum of the elements of the th row is , it follows that the sum of all elements in rows through is . The center elements are in the form , so the sum of these elements is .
The sum of the elements on or to the right of the line of symmetry is thus . However, we also counted the numbers from to . Indeed, all of these numbers have at least 's in their base- representation, as all of them are greater than , which has 's. Therefore, our answer is , and the remainder is .
Solution 2
We seek the number of allowed numbers which have 1's, not including the leading 1, for .
For , this number is
.
By the Hockey Stick Identity, this is equal to . So we get
.
For , we end on - we don't want to consider numbers with more than 11 digits. So for each we get
again by the Hockey Stick Identity. So we get
.
The total is . Subtracting out the numbers between and gives . Thus the answer is .
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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