document.write( "Question 829844: Find greatest number of six digits which on being divided by 6,7,8,9,10 leaves 4,5,6,7,8 as remainders respectively. \n" ); document.write( "
Algebra.Com's Answer #513708 by Edwin McCravy(20060)\"\" \"About 
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document.write( "Let the required number be N\r\n" );
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document.write( "then there exist positive integers A,B,C,D,E such that\r\n" );
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document.write( "N = 6A+4 = 7B+5 = 8C+6 = 9D+7 = 10E+8\r\n" );
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document.write( "Add 2 to each one:\r\n" );
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document.write( "N+2 = 6A+6 = 7B+7 = 8C+8 = 9D+9 = 10E+10\r\n" );
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document.write( "N+2 = 6(A+1) = 7(B+1) = 8(C+1) = 9(D+1) = 10(E+1)\r\n" );
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document.write( "So N+2 must be divisible by 6,7,8,9, and 10\r\n" );
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document.write( "Therefore N+2 must be a multiple of the least common multiple\r\n" );
document.write( "of 6,7,8,9,10\r\n" );
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document.write( "6=2×3\r\n" );
document.write( "7=7\r\n" );
document.write( "8=2×2×2\r\n" );
document.write( "9=3×3\r\n" );
document.write( "10=2×5\r\n" );
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document.write( "So the least common multiple of those is 2×2×2×3×3×5×7 = 2520\r\n" );
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document.write( "So N+2 must be a multiple of 2520\r\n" );
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document.write( "Therefore \r\n" );
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document.write( "N+2 = 2520k\r\n" );
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document.write( "N = 2520k-2\r\n" );
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document.write( "N is a 6 digit number, which means\r\n" );
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document.write( "100000 <= N <= 999999\r\n" );
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document.write( "100000 ≦ 2520k-2 ≦ 999999\r\n" );
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document.write( "100002 ≦ 2520k ≦ 1000001\r\n" );
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document.write( "We divide through by 2520\r\n" );
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document.write( "39.68333333... ≦ k ≦ 396.8257937...\r\n" );
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document.write( "So  40 ≦ k ≦ 396\r\n" );
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document.write( "Since we want N to be large as possible,\r\n" );
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document.write( "we take k = 396\r\n" );
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document.write( "So N = 2520(396)-2 = 997918 \r\n" );
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document.write( "Answer: 997918\r\n" );
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document.write( "Edwin
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