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

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== Solution ==
 
== Solution ==
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Let <math>(x,y)</math> denote the square in row <math>x \ge 1</math> and column <math>y \ge 1</math>. Under the first ordering this square would have a value of <math>17(x-1) + y</math>. Under the second ordering this square would have a value of <math>13(y-1) + x</math>. Equating, <math>17x-17 + y = 13y-13+x \Longrightarrow 16x = 12y + 4 \Longrightarrow 4x = 3y + 1</math>. The pairs that fit this equation are <math>(1,1),(4,5),(7,9),(10,13),(13,17)</math>; their corresponding values sum up to <math>1 + 56 + 111 + 166 + 221 = 555\ \mathrm{(D)}</math>.
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Index the rows with <math>i = 1, 2, 3, ..., 13</math>
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Index the columns with <math>j = 1, 2, 3, ..., 17</math>
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For the first row number the cells <math>1, 2, 3, ..., 17</math>
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For the second, <math>18, 19, ..., 34</math>
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and so on
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So the number in row = <math>i</math> and column = <math>j</math> is
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<math>f(i, j) = 17(i-1) + j = 17i + j - 17</math>
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 +
Similarly, numbering the same cells columnwise we
 +
find the number in row = <math>i</math> and column = <math>j</math> is
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<math>g(i, j) = i + 13j - 13</math>
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 +
So we need to solve
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<math>f(i, j) = g(i, j)</math>
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<math>17i + j - 17 = i + 13j - 13</math>
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<math>16i = 4 + 12j</math>
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<math>4i = 1 + 3j</math>
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<math>i = (1 + 3j)/4</math>
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 +
We get
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<math>(i, j) = (1, 1), f(i, j) = g(i, j) = 1</math>
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<math>(i, j) = (4, 5), f(i, j) = g(i, j) = 56</math>
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<math>(i, j) = (7, 9), f(i, j) = g(i, j) = 111</math>
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<math>(i, j) = (10, 13), f(i, j) = g(i, j) = 166</math>
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<math>(i, j) = (13, 17), f(i, j) = g(i, j) = 221</math>
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<math>\boxed{D}</math> <math>555</math>
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== Video Solution ==
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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 (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

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