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

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== Solution ==
 
== Solution ==
Index the rows with i = 1, 2, 3, ..., 13
+
Index the rows with <math>i = 1, 2, 3, ..., 13</math>
Index the columns with j = 1, 2, 3, ..., 17
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Index the columns with <math>j = 1, 2, 3, ..., 17</math>
  
For the first row number the cells 1, 2, 3, ..., 17
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For the first row number the cells <math>1, 2, 3, ..., 17</math>
For the second, 18, 19, ..., 34
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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)</math>
17i + j - 17 = i + 13j - 13
+
 
16i = 4 + 12j
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<math>17i + j - 17 = i + 13j - 13</math>
4i = 1 + 3j
+
 
i = (1 + 3j)/4
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<math>16i = 4 + 12j</math>
 +
 
 +
<math>4i = 1 + 3j</math>
 +
 
 +
<math>i = (1 + 3j)/4</math>
  
 
We get
 
We get
(i, j) = (1, 1), f(i, j) = g(i, j) = 1
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<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
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<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>(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>
  
D. 555
+
<math>\boxed{D}</math> <math>555</math>
  
Courtesy the experts on yahoo answers.
+
== Video Solution ==
 +
https://youtu.be/qCkyf2XVJcg
  
 
== See also ==
 
== See also ==

Latest revision as of 10:28, 3 August 2021

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

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

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

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

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

$\boxed{D}$ $555$

Video Solution

https://youtu.be/qCkyf2XVJcg

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 12 Problems and Solutions

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