document.write( "Question 1040980: When Joyce counts the pennies in her bank by fives, she has one left over. When she counts them by threes, there are two left over. What is the least possible number of pennies in the bank? \n" ); document.write( "

Algebra.Com's Answer #655904 by Edwin McCravy(20062) $\"\"$ $\"About$

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When Joyce counts the pennies in her bank by fives, she has one left over.

So she has 5N+1 pennies, where N is some positive integer

When she counts them by threes, there are two left over.

So she has 3M+2 pennies, where M is some other positive integer.\r\n" );
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document.write( "So 3M+2 = 5N+1 \r\n" );
document.write( "   3M-5M = -1\r\n" );
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document.write( "Write 5 in terms of its nearest multiple of 3, which is 6\r\n" );
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document.write( "   3M-(6-1)N = -1\r\n" );
document.write( "     3M-6N+N = -1\r\n" );
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document.write( "Divide through by 3\r\n" );
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document.write( "      M-2N+N/3 = -1/3\r\n" );
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document.write( "Get fractions left, whole numbers right:\r\n" );
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document.write( "       N/3+1/3 = 2N-M\r\n" );
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document.write( "Right side is a positive integer, so the right side is too.\r\n" );
document.write( "Let that positive integer by A\r\n" );
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document.write( "N/3+1/3 = A;      2N-M = A\r\n" );
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document.write( "   N+1 = 3A;      \r\n" );
document.write( "     N = 3A-1;   \r\n" );
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document.write( "Substitute in 2N-M = A\r\n" );
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document.write( "2(3A-1)-M = A\r\n" );
document.write( "   6A-2-M = A\r\n" );
document.write( "       5A = M+2\r\n" );
document.write( "     5A-2 = M\r\n" );
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document.write( "So N = 3A-1 and M = 5A-2

What is the least possible number of pennies in the bank?

The least possible value of A to make N the\r\n" );
document.write( "least positive integer is A=1.\r\n" );
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document.write( "So N = 3A-1 = 3(1)-1 = 3-1 = 2 and M = 5A-2 = 5(1)-2 = 5-2 = 3\r\n" );
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document.write( "So she has 5N+1 pennies = 5(2)+1 = 10+1 = 11 pennies\r\n" );
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document.write( "As a check:\r\n" );
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document.write( "She also has 3M+2 pennies = 3(3)+2 = 11 pennies.\r\n" );
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document.write( "And when we divide 11 by 5, we get 1 remainder, and\r\n" );
document.write( "when we divide 11 by 3, we get 2 remainder.\r\n" );
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document.write( "So 11 pennies is correct.\r\n" );
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document.write( "Edwin

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