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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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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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== See also ==
 
== See also ==

Revision as of 17:54, 6 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

The very first cell in the grid, the middle cell, and the last cell fit the criteria. These have a value of $1, 111,$ and $221$, the sum of which = $333\ \mathrm{(B)}$.


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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