document.write( "Question 355666: A woman with a basket of eggs finds that if she removes the eggs from the basket either two, three, four, five or six at a time, there is always one egg left. However, if she removes the eggs seven at a time, there are no eggs left. If the basket holds at least 300 eggs, how many eggs are in the womans basket?
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Algebra.Com's Answer #254101 by Edwin McCravy(20055)\"\" \"About 
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A woman with a basket of eggs finds that if she removes the eggs from the basket either two, three, four, five or six at a time, there is always one egg left.
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document.write( "Stop there for a minute.\r\n" );
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document.write( "To be a multiple of 2, 3, 4, 5, and 6 an integer must contain the factors\r\n" );
document.write( "2*2*3*5 or 60.  So in order for a positive integer to leave a remainder of 1\r\n" );
document.write( "when divided by 2,3,4,5 or 6 it must be 1 more than a multiple of 60, or an\r\n" );
document.write( "integer of the form 60p+1 where p is a positive integer.  Now we read the\r\n" );
document.write( "rest:       \r\n" );
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document.write( "So it must also be a multiple of 7, or 7q where q is a positive integer, so\r\n" );
document.write( "we have the linear Diophantine equation\r\n" );
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document.write( "            60p + 1 = 7q\r\n" );
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document.write( "Write all numbers in terms of their nearest multiple of the smallest\r\n" );
document.write( "coefficient in absolute value, which is 7.  So we write 60 as 63 - 3,\r\n" );
document.write( "and leave 1 as it is because the nearest multiple of 7 to 1 is 0.\r\n" );
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document.write( "      (63 - 3)p + 1 = 7q \r\n" );
document.write( "       63p - 3p + 1 = 7q\r\n" );
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document.write( "Divide through by 7\r\n" );
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document.write( "     9p - 3p/7 + 1/7 = q\r\n" );
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document.write( "Isolate fractions:\r\n" );
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document.write( "       9p - q = 3p/7 - 1/7\r\n" );
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document.write( "The right side is positive, and the left side is an integer,\r\n" );
document.write( "so both sides equal a positive integer say A, so\r\n" );
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document.write( "9p - q = A and 3p/7 - 1/7 = A\r\n" );
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document.write( "Clear the second of fractions:\r\n" );
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document.write( "3p - 1 = 7A\r\n" );
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document.write( "Write all numbers in terms of their nearest multiple of the smallest\r\n" );
document.write( "coefficient in absolute value, which is 3.  So we write 7 as 6 + 1,\r\n" );
document.write( "and leave 1 as it is because the nearest multiple of 3 to 1 is 0.\r\n" );
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document.write( "3p - 1 = (6 + 1)A\r\n" );
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document.write( "3p - 1 = 6A + A\r\n" );
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document.write( "Divide thru by 3\r\n" );
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document.write( "p - 1/3 = 2A + A/3\r\n" );
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document.write( "Isolate fractions:\r\n" );
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document.write( "p - 2A = 1/3 + A/3\r\n" );
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document.write( "The left side is an integer and the right side is positive,\r\n" );
document.write( "so both sides equal a positive integer, say B\r\n" );
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document.write( "p - 2A = B and 1/3 + A/3 = B\r\n" );
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document.write( "Clearing the second equation of fractions,\r\n" );
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document.write( "1 + A = 3B\r\n" );
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document.write( "A = 3B - 1 \r\n" );
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document.write( "Substituting in p - 2A = B\r\n" );
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document.write( "p - 2(3B - 1) = B\r\n" );
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document.write( "   p - 6B + 2 = B\r\n" );
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document.write( "            p = 7B - 2\r\n" );
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document.write( "So \r\n" );
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document.write( "60p + 1 = 60(7B - 2) + 1 = 420B - 120 + 1 = 420B - 119\r\n" );
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document.write( "Thus she can have 420B - 119 eggs in her basket, where B is any\r\n" );
document.write( "positive integer.\r\n" );
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document.write( "For B = 1, she can have 301 eggs\r\n" );
document.write( "For B = 2, she can have 721 eggs\r\n" );
document.write( "For B = 3, she can have 1141 eggs \r\n" );
document.write( "...\r\n" );
document.write( "etc., etc., etc.\r\n" );
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document.write( "Edwin
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