Question 1142439
<pre>

Let the number be  ABBBC

A can be any one of 4,5, or 6

C can only be 5, 6, 7, 8, or 9,  subject to C > A

Finally, A+3B+C = 19


We want 19-A-C to be divisible by 3, subject to C > A, there are three possible cases to consider:
19-10 = 9
19-13 = 6
19-16 = 3

So we need to check cases where A+C = 10, A+C = 13, and A+C = 16 


A+C = 10:
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A=4, C=6:  19-4-6 = 9  --->  B = 9/3 = 3,   so 43336 is one solution


A+C = 13:
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A=4, C=9: 19-4-9 = 6  --->  B = 2,  so  42229  is another solution
A=5, C=8: 19-5-8 = 6  --->  B = 2,  so  52228  is a solution
A=6, C=7: 19-6-7 = 6  --->  B = 2,  so  62227  is a solution


A+C = 16:
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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.
 

I think these are all of the solutions: 43336, 42229, 52228, and 62227