document.write( "Question 1075115: 6x=2 mod7 \n" ); document.write( "

Algebra.Com's Answer #689804 by ikleyn(52803) $\"\"$ $\"About$

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document.write( "The solution is x = 5 (mod 7).\r\n" );
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document.write( "Because 6x = 6*5 = 30 gives the remainder 2 when divided by 7.\r\n" );
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document.write( "In other words, in the ring (in the field !) of integers (mod 7) the solution to the equation\r\n" );
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document.write( "    6x = 5 (mod 7) \r\n" );
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document.write( "    x = 5 (mod 7).\r\n" );
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document.write( "Or, in the multiplicative group of integers (mod 7) the solution to the equation\r\n" );
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document.write( "    6x = 5 (mod 7) \r\n" );
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document.write( "    x = 5 (mod 7).\r\n" );
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document.write( "Or, in the additive group of integers (mod 7) the solution to the equation\r\n" );
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document.write( "    6x = 5 (mod 7) \r\n" );
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document.write( "is \r\n" );
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document.write( "    x = 5 (mod 7).\r\n" );
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