document.write( "Question 147578: In how many ways can you arrange the digits 8, 7, 6, and 3 to form a four-digit number divisible by 9, using each digit once and only once? \n" ); document.write( "

Algebra.Com's Answer #108163 by Edwin McCravy(20055) $\"\"$ $\"About$

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In how many ways can you arrange the digits 8, 7, 6, and 3 to form a four-digit number divisible by 9, using each digit once and only once?\r
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document.write( "There is no way! Why? Because the sum of the \r\n" );
document.write( "digits of any multiple of 9 must itself be a \r\n" );
document.write( "multiple of 9. However the sum of these digits \r\n" );
document.write( "is 8+7+6+3=24 which is not a multiple of 9, \r\n" );
document.write( "regardless of how you arrange them.\r\n" );
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document.write( "Proof that the sum of the digits of a multiple of\r\n" );
document.write( "9 is a multiple of 9:\r\n" );
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document.write( "Suppose the number which is a multiple of 9 \r\n" );
document.write( "has the digits A,B,C, and D.  Then the number is\r\n" );
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document.write( "1000A + 100B + 10C + D\r\n" );
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document.write( "Suppose this equals to a multiple of 9, \r\n" );
document.write( "say, 9 times an integer N, or 9N\r\n" );
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document.write( "1000A + 100B + 10C + D = 9N\r\n" );
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document.write( "We write 1000 as 999+1, 100 as 99+1, \r\n" );
document.write( "and 10 as 9+1:\r\n" );
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document.write( "(999+1)A + (99+1)B + (9+1)C + D = 9N\r\n" );
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document.write( "999A + A + 99B + B + 9C + C + D = 9N\r\n" );
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document.write( "A + B + C + D = 9N - 999A - 99B - 9C\r\n" );
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document.write( "Factor out 9 on the right:\r\n" );
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document.write( "A + B + C + D = 9(N - 111A - 11B - C)\r\n" );
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document.write( "The right side is a multiple of 9 and the \r\n" );
document.write( "left side is the sum of the digits.\r\n" );
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document.write( "That proves that the sum of the digits of \r\n" );
document.write( "a multiple of 9 is also a multiple of 9.\r\n" );
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document.write( "Therefore no matter how you arrange the \r\n" );
document.write( "digits 8, 7, 6, and 3, it can never be a \r\n" );
document.write( "multiple of 9, because their sum is 24, \r\n" );
document.write( "which is not a multiple of 9.\r\n" );
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document.write( "On the other hand, if you could trade the \r\n" );
document.write( "6 for a 9,  then the sum of the digits \r\n" );
document.write( "would be 27, which is a multiple of 9,\r\n" );
document.write( "and therefore all 24 ways to arrange \r\n" );
document.write( "the digits 8, 7, 9, and 3 will be a \r\n" );
document.write( "multiple of 9, as you see below:\r\n" );
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document.write( " 1.  3789 = 9×421\r\n" );
document.write( " 2.  3798 = 9×422\r\n" );
document.write( " 3.  3879 = 9×431\r\n" );
document.write( " 4.  3897 = 9×433\r\n" );
document.write( " 5.  3978 = 9×442\r\n" );
document.write( " 6.  3987 = 9×443\r\n" );
document.write( " 7.  7389 = 9×821\r\n" );
document.write( " 8.  7398 = 9×822\r\n" );
document.write( " 9.  7839 = 9×871\r\n" );
document.write( "10.  7893 = 9×877\r\n" );
document.write( "11.  7938 = 9×882\r\n" );
document.write( "12.  7983 = 9×887\r\n" );
document.write( "13.  8379 = 9×931\r\n" );
document.write( "14.  8397 = 9×933\r\n" );
document.write( "15.  8739 = 9×971\r\n" );
document.write( "16.  8793 = 9×977\r\n" );
document.write( "17.  8937 = 9×993\r\n" );
document.write( "18.  8973 = 9×997\r\n" );
document.write( "19.  9378 = 9×1042\r\n" );
document.write( "20.  9387 = 9×1043\r\n" );
document.write( "21.  9738 = 9×1082\r\n" );
document.write( "22.  9783 = 9×1087\r\n" );
document.write( "23.  9837 = 9×1093\r\n" );
document.write( "24.  9873 = 9×1097\r\n" );
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document.write( "Edwin

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