Difference between revisions of "2000 AMC 12 Problems/Problem 16"
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<math>f(i, j) = g(i, j) | <math>f(i, j) = g(i, j) | ||
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17i + j - 17 = i + 13j - 13 | 17i + j - 17 = i + 13j - 13 | ||
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16i = 4 + 12j | 16i = 4 + 12j | ||
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4i = 1 + 3j | 4i = 1 + 3j | ||
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i = (1 + 3j)/4</math> | i = (1 + 3j)/4</math> | ||
We get | We get | ||
<math>(i, j) = (1, 1), f(i, j) = g(i, j) = 1 | <math>(i, j) = (1, 1), f(i, j) = g(i, j) = 1 | ||
+ | |||
(i, j) = (4, 5), f(i, j) = g(i, j) = 56 | (i, j) = (4, 5), f(i, j) = g(i, j) = 56 | ||
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(i, j) = (7, 9), f(i, j) = g(i, j) = 111 | (i, j) = (7, 9), f(i, j) = g(i, j) = 111 | ||
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(i, j) = (10, 13), f(i, j) = g(i, j) = 166 | (i, j) = (10, 13), f(i, j) = g(i, j) = 166 | ||
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(i, j) = (13, 17), f(i, j) = g(i, j) = 221</math> | (i, j) = (13, 17), f(i, j) = g(i, j) = 221</math> | ||
Revision as of 19:45, 10 June 2017
Problem
A checkerboard of rows and columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered , the second row , and so on down the board. If the board is renumbered so that the left column, top to bottom, is , the second column and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).
Solution
Index the rows with Index the columns with
For the first row number the cells For the second, and so on
So the number in row = and column = is
Similarly, numbering the same cells columnwise we find the number in row = and column = is
So we need to solve
$f(i, j) = g(i, j)
17i + j - 17 = i + 13j - 13
16i = 4 + 12j
4i = 1 + 3j
i = (1 + 3j)/4$ (Error compiling LaTeX. ! Missing $ inserted.)
We get $(i, j) = (1, 1), f(i, j) = g(i, j) = 1
(i, j) = (4, 5), f(i, j) = g(i, j) = 56
(i, j) = (7, 9), f(i, j) = g(i, j) = 111
(i, j) = (10, 13), f(i, j) = g(i, j) = 166
(i, j) = (13, 17), f(i, j) = g(i, j) = 221$ (Error compiling LaTeX. ! Missing $ inserted.)
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.