document.write( "Question 416972: coin collector had 6 coins for sale
\n" ); document.write( "one he price @15,16,18,19,20,31 he sold all but one of the coins
\n" ); document.write( "to 2 people one payinng 2xs as much as the other one
\n" ); document.write( "which coin did he keep
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Algebra.Com's Answer #292019 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
I just tried every single number that could be the one left out
\n" ); document.write( "First I let \"x\" = the amount one person paid and \"2x\" =
\n" ); document.write( "amount other person paid. \"x+%2B+2x+=+3x\", so the sum of
\n" ); document.write( "what they bought must be divisible by \"3\"
\n" ); document.write( "The only way I could arrive at this is if the coin costing 20
\n" ); document.write( "is left out. Then \"15+%2B+16+%2B+18+%2B+19+%2B+31+=+99\", so if
\n" ); document.write( "\"x+%2B+2x+=+99\"
\n" ); document.write( "\"3x+=+99\"
\n" ); document.write( "\"x+=+33\"
\n" ); document.write( "1 person bought 33 worth of coins
\n" ); document.write( "The other bought 66 worth
\n" ); document.write( "\"15+%2B+18+=+33\"
\n" ); document.write( "and
\n" ); document.write( "\"+16+%2B+19+%2B+31+=+66+\"
\n" ); document.write( "So, he kept the coin costing 20
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