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

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== Solution ==
 
== Solution ==
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The very first cell in the grid, the middle cell, and the last cell fit the criteria.  These have a value of <math>1, 111,</math> and <math>221</math>, the sum of which = <math>333\ \mathrm{(B)}</math>.
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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, 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)
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17i + j - 17 = i + 13j - 13
 +
16i = 4 + 12j
 +
4i = 1 + 3j
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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
 +
D. 555
  
 
== See also ==
 
== See also ==

Revision as of 20:49, 14 January 2015

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, 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 D. 555

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. AMC logo.png

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, 18, 19, ..., 34 and so on

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 D. 555

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