document.write( "Question 1108009: The five-digit number 9abc2 is divisible by 11, where a,b,and c are different one-digit whole numbers. How many combinations of digits are possible values of a,b,and c? \n" ); document.write( "
Algebra.Com's Answer #723192 by Edwin McCravy(20056) $\"\"$ $\"About$
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The above solution is incorrect because it ignores the restriction\r\n" );
document.write( "that a, b, and c are DIFFERENT one-digit whole numbers.  Here is the\r\n" );
document.write( "correct solution:\r\n" );
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document.write( "Note:  I will use the rule followed in \" 9abc2 \". That is, if no multiplication\r\n" );
document.write( "symbol is placed between either two letters or between a digit and a letter,\r\n" );
document.write( "then the letter is assumed to be a digit and no multiplication symbol is ever\r\n" );
document.write( "understood and not written.  Multiplication will only be assumed if the symbol\r\n" );
document.write( "\"×\" appears.\r\n" );
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document.write( "9abc2 = 90002 + abc0\r\n" );
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document.write( "Since 90002 is divisible by 11, [90002 = 11×8182]\r\n" );
document.write( "9abc2 will be divisible by 11 if and only if abc0 \r\n" );
document.write( "is also divisible by 11.\r\n" );
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document.write( "Since abc0 = 10×abc, abc0 will be divisible by\r\n" );
document.write( "11 if and only if abc is divisible by 11.\r\n" );
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document.write( "Therefore if abc is any term of this arithmetic sequence\r\n" );
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document.write( "S = 000, 011,022,...,099,110,121,132,...,968,979,990, \r\n" );
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document.write( "then 9abc2 will be divisible by 11\r\n" );
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document.write( "If we divide every term of sequence S by 11, we\r\n" );
document.write( "get the arithmetic sequence:\r\n" );
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document.write( "0,1,2,...,9,10,11,12,...,88,89,90\r\n" );
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document.write( "That's 91 terms, since it starts with 0 and ends with 90.\r\n" );
document.write( "So if we did not have the requirement that a,b, and c\r\n" );
document.write( "must all three be different digits, the answer would be 91\r\n" );
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document.write( "[Note: this 91 is what the other tutor gave as a final solution.]\r\n" );
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document.write( "But from the 91 terms of multiples of 11 (less than 1000),\r\n" );
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document.write( "we must eliminate all that have 2 or three digits the same:\r\n" );
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document.write( "Case 1: a=b, that is abc = aac\r\n" );
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document.write( "then\r\n" );
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document.write( "abc = 100×a+10×a+c\r\n" );
document.write( "abc = 110×a + c\r\n" );
document.write( "abc-110×a = c\r\n" );
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document.write( "Both terms on the left are divisible by 11, so c must be\r\n" );
document.write( "a digit divisible by 11, and 0 is the only digit divisible by\r\n" );
document.write( "11.   Thus c=0,\r\n" );
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document.write( "So we must remove this arithmetic sequence:\r\n" );
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document.write( "000,110,220,...,990\r\n" );
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document.write( "If we divide every term of that arithmetic sequence by 11, we\r\n" );
document.write( "get the arithmetic sequence \r\n" );
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document.write( "0,1,2,...,9\r\n" );
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document.write( "That's 10 terms, since it starts with 0 and ends with 9.\r\n" );
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document.write( "Thus by case 1, we remove 10 terms from sequence S.\r\n" );
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document.write( "Case 2: b=c, that is abc = abb\r\n" );
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document.write( "then\r\n" );
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document.write( "abc = 100×a+10×b+b\r\n" );
document.write( "abc = 100×a + 11×b\r\n" );
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document.write( "Since abc is divisible by 11, there exists integer K such that abc = 11×K\r\n" );
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document.write( "11×K = 100×a + 11×b\r\n" );
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document.write( "11×K - 11×b = 100×a\r\n" );
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document.write( "Both terms on the left are divisible by 11, and 100 is not divisible by 11,\r\n" );
document.write( "so a must be a digit divisible by 11, and 0 is the only digit divisible by\r\n" );
document.write( "11.   Thus a=0,\r\n" );
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document.write( "So we must remove this arithmetic sequence:\r\n" );
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document.write( "000,011,022,...,099\r\n" );
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document.write( "We have already removed 000 in case 1, so in case 2 we need remove only\r\n" );
document.write( "the other 9 terms.\r\n" );
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document.write( "Thus for case 2, we remove 9 terms from sequence S.\r\n" );
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document.write( "Case 3: a=c, that is abc = aba  [The most difficult case].\r\n" );
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document.write( "then\r\n" );
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document.write( "abc = 100×a+10×b+a\r\n" );
document.write( "abc = 101×a + 10×b\r\n" );
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document.write( "Since abc is divisible by 11, there exists integer K such that abc = 11×K\r\n" );
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document.write( "11×K = 101×a + 10×b\r\n" );
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document.write( "We write 101×a as 121×a - 22×a +2×a, and 10×b as 11×b-b\r\n" );
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document.write( "11×K = 121×a - 22×a +2×a + 11×b-b\r\n" );
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document.write( "We divide through by 11\r\n" );
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document.write( "K = 11×a - 2×a  + (2/11)×a + b - c/11\r\n" );
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document.write( "We isolate the fractions on the left:\r\n" );
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document.write( "c/11 - (2/11)×a = 11×a - 2×a + b - K\r\n" );
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document.write( "The right side is an integer, so the left side must also be an integer,\r\n" );
document.write( "say the integer P\r\n" );
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document.write( "c/11 - (2/11)×a = P\r\n" );
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document.write( "Multiplying through by 11\r\n" );
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document.write( "c - 2×a = 11×P\r\n" );
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document.write( "c = 2×a + 11×P  <-- this could also be written c = (2×a) mod 11\r\n" );
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document.write( "P could only be 0 or -1, for no other values will allow c to be a digit.\r\n" );
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document.write( "Sub-case 3A: P = 0\r\n" );
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document.write( "c = 2×a,  this gives only possibilities a=0,1,2,3,4 and we have already\r\n" );
document.write( "counted 000 in case 1.\r\n" );
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document.write( "Thus for sub-case 3A, we remove 4 more terms from sequence S.\r\n" );
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document.write( "[FYI, the 4 terms removed from S here are 121, 242, 363, 484]\r\n" );
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document.write( "Sub-case 3B: P = -1\r\n" );
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document.write( "c = 2×a-11,  this gives only possibilities a=6,7,8,9.\r\n" );
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document.write( "Thus for sub-case 3B, we remove 4 more terms from sequence S.\r\n" );
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document.write( "[FYI, the 4 terms removed from S here are 616, 737, 858, 979]\r\n" );
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document.write( "Thus for case 3, we remove 4+4=8  terms from sequence S.\r\n" );
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document.write( "So the final answer is 91-10-9-8 = 64.\r\n" );
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document.write( "Edwin
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