document.write( "Question 1084278: The total number of subtractions that result in 11111 remaining after a four-digit number has been subtracted from a five-digit 1 through 9 have all been used is \n" ); document.write( "

Algebra.Com's Answer #698467 by math_helper(2461) $\"\"$ $\"About$

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The five digit number is $\"+a%2A10%5E4+%2B+b%2A10%5E3+%2B+c%2A10%5E2+%2B+d%2A10+%2B+e+\"$
\n" ); document.write( "and the four digit number is $\"++f%2A10%5E3+%2B+g%2A10%5E2+%2B+h%2A10+%2B+j+\"$ \r
\n" ); document.write( "\n" ); document.write( "We have:\r
\n" ); document.write( "\n" ); document.write( "b-f = 1
\n" ); document.write( "c-g = 1
\n" ); document.write( "d-h = 1
\n" ); document.write( "e-j = 1
\n" ); document.write( "and a = 1\r
\n" ); document.write( "\n" ); document.write( "Each digit of the 5-digit number is one more than the corresponding digit in the 4-digit number, and the \"1\" is fixed at the start of the 5-digit number. \r
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\n" ); document.write( "\n" ); document.write( "Since all the digits 1…9 are used, the numbers must make use of the digits:
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\n" ); document.write( " 13579 and 2468, respectively \r
\n" ); document.write( "\n" ); document.write( "The \"1\" is fixed, but the 3,5,7, and 9 can be arranged in any order (that's 4! ways = 24 ways), and the 2,4,6,8 are simply re-arranged to follow along (placing 2 below 3, 6 below 7, etc., in the subtraction). \r
\n" ); document.write( "\n" ); document.write( "So $\"+highlight%2824%29+\"$ subtractions of a 4-digit number from a 5-digit number exist, where 1…9 are all used. \r
\n" ); document.write( "\n" ); document.write( "While I have not proved it rigorously, no other arrangements will give you 11111.\r
\n" ); document.write( "\n" ); document.write( "—
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document.write( "A few examples:\r\n" );
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document.write( "13579\r\n" );
document.write( "-2468\r\n" );
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document.write( "19753\r\n" );
document.write( "-8642\r\n" );
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document.write( "19375\r\n" );
document.write( "-8264\r\n" );
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document.write( "etc.\r\n" );
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