document.write( "Question 965781: From 1,2,3,4,5 and 6 form a four digit number (where repetition is not allowed). How many such numbers are divisible by 3? \n" ); document.write( "

Algebra.Com's Answer #590296 by Edwin McCravy(20055) $\"\"$ $\"About$

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From 1,2,3,4,5 and 6 form a four digit number (where repetition is not allowed). How many such numbers are divisible by 3?
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document.write( "In order to be divisible by 3, the sum of its digits must be divisible by 3.\r\n" );
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document.write( "The sum of the digits of any such 4-digit number cannot be less than\r\n" );
document.write( "1+2+3+4=10 nor greater than 3+4+5+6=18.\r\n" );
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document.write( "Therefore the sum of the digits must be 12, 15 or 18, the only sums of digits\r\n" );
document.write( "divisible by 3 in that range.\r\n" );
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document.write( "Case 1.  The number has sum of digits 12.\r\n" );
document.write( "To have sum of digits 12, the digits must be an arrangement of 1,2,4,5 or\r\n" );
document.write( "1,2,3,6\r\n" );
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document.write( "Here's why:\r\n" );
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document.write( "  a. If 1 were omitted, the smallest possible sum would be 2+3+4+5=14.  Too\r\n" );
document.write( "     large\r\n" );
document.write( "  b. Since you must have a 1, if you leave out 2 then the smallest possible sum\r\n" );
document.write( "     would be 1+3+4+5=13.  Still too large.\r\n" );
document.write( "  c. Since 1+2=3, the other two digits must have sum 9, which is only possible\r\n" );
document.write( "     with 4+5 or 6+3 \r\n" );
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document.write( "So there are 4! permutations of the digits 1,2,4,5 and also\r\n" );
document.write( "   there are 4! permutations of the digits 1,2,3,6.\r\n" );
document.write( "So that 2*4! = 2*24 = 48 with sum of digits 12. \r\n" );
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document.write( "Case 2.  The number has sum of digits 15.\r\n" );
document.write( "To have sum of digits 15, the digits must be an arrangement of 2,3,4,6 or\r\n" );
document.write( "1,3,5,6.\r\n" );
document.write( "  \r\n" );
document.write( "Here's why:\r\n" );
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document.write( "  a. It must have digit 6, for if 6 were omitted then the greatest possible sum\r\n" );
document.write( "     of digits would be only 2+3+4+5=14. So the number must have a 6 as a digit.\r\n" );
document.write( "  \r\n" );
document.write( "  b. Since it contains a digit 6, the other 3 digits must have sum 9 and\r\n" );
document.write( "     selected from 1,2,3,4,5.\r\n" );
document.write( "     \r\n" );
document.write( "  c. A quick inspection shows that 1+3+5=9 and 2+3+4=9 are the only\r\n" );
document.write( "     possibilities for a sum of the remaining 3 digits to be 9. \r\n" );
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document.write( "So there are 4! permutations of the digits 2,3,4,6 and also\r\n" );
document.write( "there are 4! permutations of the digits 1,3,5,6.\r\n" );
document.write( "So that 2*4! = 2*24 = 48 with sum of digits 15. \r\n" );
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document.write( "Case 3.  The number has sum of digits 18. \r\n" );
document.write( "Only one combination of 4 digits sums to 18, 3+4+5+6\r\n" );
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document.write( "So there are 4! permutations of the digits 2,3,4,6.\r\n" );
document.write( "That's 4! = 24 for case 3.\r\n" );
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document.write( "Answer: 48+48+24 = 120\r\n" );
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document.write( "Edwin

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