Difference between revisions of "2000 AMC 12 Problems/Problem 16"

(Solution)
(Problem)
Line 1: Line 1:
== Problem ==
 
A checkerboard of <math>13</math> rows and <math>17</math> columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered <math>1,2,\ldots,17</math>, the second row <math>18,19,\ldots,34</math>, and so on down the board. If the board is renumbered so that the left column, top to bottom, is <math>1,2,\ldots,13,</math>, the second column <math>14,15,\ldots,26</math> 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).
 
 
<math>\text {(A)}\ 222 \qquad \text {(B)}\ 333\qquad \text {(C)}\ 444 \qquad \text {(D)}\ 555 \qquad \text {(E)}\ 666</math>
 
 
 
== Solution ==
 
== Solution ==
 
Index the rows with <math>i = 1, 2, 3, ..., 13</math>
 
Index the rows with <math>i = 1, 2, 3, ..., 13</math>

Revision as of 00:36, 10 October 2020

Solution

Index the rows with $i = 1, 2, 3, ..., 13$ Index the columns with $j = 1, 2, 3, ..., 17$

For the first row number the cells $1, 2, 3, ..., 17$ For the second, $18, 19, ..., 34$ and so on

So the number in row = $i$ and column = $j$ is $f(i, j) = 17(i-1) + j = 17i + j - 17$

Similarly, numbering the same cells columnwise we find the number in row = $i$ and column = $j$ is $g(i, j) = i + 13j - 13$

So we need to solve

$f(i, j) = g(i, j)$

$17i + j - 17 = i + 13j - 13$

$16i = 4 + 12j$

$4i = 1 + 3j$

$i = (1 + 3j)/4$

We get $(i, j) = (1, 1), f(i, j) = g(i, j) = 1$

$(i, j) = (4, 5), f(i, j) = g(i, j) = 56$

$(i, j) = (7, 9), f(i, j) = g(i, j) = 111$

$(i, j) = (10, 13), f(i, j) = g(i, j) = 166$

$(i, j) = (13, 17), f(i, j) = g(i, j) = 221$

$\boxed{D}$ $555$

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
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. AMC logo.png