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

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== Solution ==
 
== Solution ==
 
Index the rows with <math>i = 1, 2, 3, ..., 13</math>
 
Index the rows with <math>i = 1, 2, 3, ..., 13</math>
Index the columns with j = 1, 2, 3, ..., 17
+
Index the columns with <math>j = 1, 2, 3, ..., 17</math>
  
For the first row number the cells 1, 2, 3, ..., 17
+
For the first row number the cells <math>1, 2, 3, ..., 17</math>
For the second, 18, 19, ..., 34
+
For the second, <math>18, 19, ..., 34</math>
 
and so on
 
and so on
  
So the number in row = i and column = j is
+
So the number in row = <math>i</math> and column = <math>j</math> is
f(i, j) = 17(i-1) + j = 17i + j - 17
+
<math>f(i, j) = 17(i-1) + j = 17i + j - 17</math>
  
 
Similarly, numbering the same cells columnwise we
 
Similarly, numbering the same cells columnwise we
find the number in row = i and column = j is
+
find the number in row = <math>i</math> and column = <math>j</math> is
g(i, j) = i + 13j - 13
+
<math>g(i, j) = i + 13j - 13</math>
  
 
So we need to solve
 
So we need to solve
  
f(i, j) = g(i, j)
+
<math>f(i, j) = g(i, j)
 
17i + j - 17 = i + 13j - 13
 
17i + j - 17 = i + 13j - 13
16i = 4 + 12j
+
16i = 4 + 12j  
 
4i = 1 + 3j
 
4i = 1 + 3j
i = (1 + 3j)/4
+
i = (1 + 3j)/4</math>
  
 
We get
 
We get
(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
 
(i, j) = (7, 9), f(i, j) = g(i, j) = 111
 
(i, j) = (7, 9), f(i, j) = g(i, j) = 111
 
(i, j) = (10, 13), f(i, j) = g(i, j) = 166
 
(i, j) = (10, 13), f(i, j) = g(i, j) = 166
(i, j) = (13, 17), f(i, j) = g(i, j) = 221  
+
(i, j) = (13, 17), f(i, j) = g(i, j) = 221</math>
  
<math>\boxed{D}</math> 555
+
<math>\boxed{D}</math> <math>555</math>
  
 
== See also ==
 
== See also ==

Revision as of 06:33, 14 October 2016

Problem

A checkerboard of $13$ rows and $17$ columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered $1,2,\ldots,17$, the second row $18,19,\ldots,34$, and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1,2,\ldots,13,$, the second column $14,15,\ldots,26$ 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).

$\text {(A)}\ 222 \qquad \text {(B)}\ 333\qquad \text {(C)}\ 444 \qquad \text {(D)}\ 555 \qquad \text {(E)}\ 666$

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

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