Difference between revisions of "2000 AMC 12 Problems/Problem 16"
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So we need to solve | So we need to solve | ||
| − | <math>f(i, j) = g(i, j) | + | <math>f(i, j) = g(i, j)</math> |
| − | 17i + j - 17 = i + 13j - 13 | + | |
| − | 16i = 4 + 12j | + | <math>17i + j - 17 = i + 13j - 13</math> |
| − | 4i = 1 + 3j | + | |
| − | i = (1 + 3j)/4</math> | + | <math>16i = 4 + 12j</math> |
| + | |||
| + | <math>4i = 1 + 3j</math> | ||
| + | |||
| + | <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</math> |
| − | (i, j) = (4, 5), f(i, j) = g(i, j) = 56 | + | |
| − | (i, j) = (7, 9), f(i, j) = g(i, j) = 111 | + | <math>(i, j) = (4, 5), f(i, j) = g(i, j) = 56</math> |
| − | (i, j) = (10, 13), f(i, j) = g(i, j) = 166 | + | |
| − | (i, j) = (13, 17), f(i, j) = g(i, j) = 221</math> | + | <math>(i, j) = (7, 9), f(i, j) = g(i, j) = 111</math> |
| + | |||
| + | <math>(i, j) = (10, 13), f(i, j) = g(i, j) = 166</math> | ||
| + | |||
| + | <math>(i, j) = (13, 17), f(i, j) = g(i, j) = 221</math> | ||
<math>\boxed{D}</math> <math>555</math> | <math>\boxed{D}</math> <math>555</math> | ||
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/qCkyf2XVJcg | ||
== See also == | == See also == | ||
Latest revision as of 11:28, 3 August 2021
Contents
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
We get
Video Solution
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. 
