document.write( "Question 1142439: am five-digit number greater than 40,000 but less than 70,000. My ones digit is more than my ten thousands digit. All my other digits are the same. The sum of my digits is 19. What number am I? \n" ); document.write( "
Algebra.Com's Answer #763136 by math_helper(2461)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Let the number be  ABBBC\r
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document.write( "A can be any one of 4,5, or 6\r
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document.write( "C can only be 5, 6, 7, 8, or 9,  subject to C > A\r
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document.write( "Finally, A+3B+C = 19\r
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document.write( "We want 19-A-C to be divisible by 3, subject to C > A, there are three possible cases to consider:
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document.write( "19-10 = 9
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document.write( "19-13 = 6
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document.write( "19-16 = 3\r
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document.write( "So we need to check cases where A+C = 10, A+C = 13, and A+C = 16 \r
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document.write( "A+C = 10:
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document.write( "A=4, C=6:  19-4-6 = 9  --->  B = 9/3 = 3,   so 43336 is one solution\r
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document.write( "A+C = 13:
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document.write( "A=4, C=9: 19-4-9 = 6  --->  B = 2,  so  42229  is another solution
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document.write( "A=5, C=8: 19-5-8 = 6  --->  B = 2,  so  52228  is a solution
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document.write( "A=6, C=7: 19-6-7 = 6  --->  B = 2,  so  62227  is a solution\r
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document.write( "A+C = 16:
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document.write( "There are no cases for A+C = 16, for even if we choose the maximum value of A, which is A=6,  we find C=10 and that is not a single digit.
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document.write( "I think these are all of the solutions: 43336, 42229, 52228, and 62227\r
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