Difference between revisions of "2007 AIME II Problems/Problem 13"
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− | Induction Step: Assume that the statement is true for n, and we wish to prove it for <math>n + 1</math>. Therefore, we add another row below the <math>n + 1</math> row, and name the squares as <math>a_0, a_1, \cdots a_{n+1}</math>. We obtain that <math>x_0 = a_0 + a_1, x_1 = a_1 + a_2, \cdots x_{n} = a_{n} + a_{n+1}</math>, and using the assumption, yield the top square equals <math>{n\choose0}(a_0 + a_1) + {n\choose1}(a_1 + a_2) + \ldots {n\choose{n}}(a_{n} + a_{n+1})</math>. While it's easy to prove that <math>{n\choose{k}} + {n\choose{k+1}} = {n+1\choose{k+1}</math> (just expand it), we find that the two equations are identical. | + | Induction Step: Assume that the statement is true for n, and we wish to prove it for <math>n + 1</math>. Therefore, we add another row below the <math>n + 1</math> row, and name the squares as <math>a_0, a_1, \cdots a_{n+1}</math>. We obtain that <math>x_0 = a_0 + a_1, x_1 = a_1 + a_2, \cdots x_{n} = a_{n} + a_{n+1}</math>, and using the assumption, yield the top square equals <math>{n\choose0}(a_0 + a_1) + {n\choose1}(a_1 + a_2) + \ldots {n\choose{n}}(a_{n} + a_{n+1})</math>. While it's easy to prove that <math>{n\choose{k}} + {n\choose{k+1}} = {n+1\choose{k+1}}</math> (just expand it), we find that the two equations are identical. |
Note contributed by MVP_Harry, friend me if you want ~ | Note contributed by MVP_Harry, friend me if you want ~ |
Revision as of 15:03, 2 January 2020
Contents
Problem
A triangular array of squares has one square in the first row, two in the second, and in general, squares in the th row for With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in given diagram). In each square of the eleventh row, a or a is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of 's and 's in the bottom row is the number in the top square a multiple of ?
Solution
Label each of the bottom squares as .
Through induction, we can find that the top square is equal to . (This also makes sense based on a combinatorial argument: the number of ways a number can "travel" to the top position going only up is equal to the number of times it will be counted in the final sum.)
Examine the equation . All of the coefficients from will be multiples of (since the numerator will have a ). Thus, the expression boils down to . Reduce to find that . Out of , either all are equal to , or three of them are equal to . This gives possible combinations of numbers that work.
The seven terms from can assume either or , giving us possibilities. The answer is therefore .
Note
The specific induction process: we want to prove that if the bottom row is , the top square is .
Base Case: when n = 1, this is obviously true.
Induction Step: Assume that the statement is true for n, and we wish to prove it for . Therefore, we add another row below the row, and name the squares as . We obtain that , and using the assumption, yield the top square equals . While it's easy to prove that (just expand it), we find that the two equations are identical.
Note contributed by MVP_Harry, friend me if you want ~
See also
2007 AIME II (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.