document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1170573>Question 1170573</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Dinner guests are to be seated at 10 round tables labeled from A to J. Table A has 21 seats numbered consecutively from 1 to 21, table B has 22 seats numbered similarly, table C has 23 numbered seats, and so on to table J having 30 seats. At each table, starting to count at seat 1, every third person will be eliminated. Here's an example:<BR>\n" );
document.write( "At a table of 12: in the first round, persons 3, 6, 9, and 12 are gone. Then, 4, 8,and 1 are eliminated. Numbers 7 and 2 are next to go. Number 11 goes next, and then 5. The winner is number 10: (The count on the last step is 5-10-5.) You have 20 seconds to choose a seat at any table from A to J. What equation can you use so that you can make a lucrative choice in 20 seconds? (x is the number of seats at a table and y is the seat that you take.)\r<BR>\n" );
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document.write( "a) y = 3x-61<BR>\n" );
document.write( "b) y = 4x-83<BR>\n" );
document.write( "c) y = 3x-69<BR>\n" );
document.write( "d) y = 3x-57<BR>\n" );
document.write( "e) y = 3x-72  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #795515 by </B><A HREF=/tutors/aboutme.mpl?userid=akumpo><B>akumpo(8)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=akumpo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=akumpo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=795515\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It would be very to type the real solution.\r<BR>\n" );
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document.write( "You can actually remove some of the options using process of elimination. y must be a positive, as there cannot be a negative seat number. x is equal to the number of seats at the table. Let's start off with the first table, which has 21 seats. 21*3=63, and 21*4=84. We can eliminate options c and e, as they would give us a negative number as our seat. \r<BR>\n" );
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document.write( "Now for the solution. Table A has 21 seats. Every 3rd person is eliminated. 3,6,9,12,15,18, and 21 will all be eliminated. Any equation that gives us a multiple of 3 as our answer cannot be true. Therefore, we can also eliminate option D as 3*21-57 would give us 6, which is a multiple of 3.\r<BR>\n" );
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document.write( "This leaves options A and B. Since they both satisfy table A, let's move to table B. Table B has 22 people, and we know y cannot equal a multiple of 3 or a negative number. 22*3-61=5, and 22*4-83=5 as well. They both satisfy table B, so move on to table C, which has 23 seats. 23*3-61=8, but 23*4-83=9.\r<BR>\n" );
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document.write( "Option B would give us a multiple of 3, so B cannot be the correct answer.<BR>\n" );
document.write( "This leaves option A as the only possible answer. You will find that the equation in option A will never give a multiple of 3 for numbers 21-30.\r<BR>\n" );
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document.write( "So, y=3x-61 is the correct answer. You can solve this without using process of elimination as well, but that would be hard to type out.\r<BR>\n" );
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document.write( "Edit: If you simply cross out every third person from 1 to 21 and repeating the process until you get one number left, seat number 2 would win for table A, number 5 for table B, number 8 for table C, etc. </SPAN>\n" );
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