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:

=========

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





A+C = 13:

=========

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:

=========

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







 

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
 The number is

N x x x M

M > N

M is either 4, 5, or 6

N + M + 3x = 19

x cannot be 9

x cannot be 8

x cannot be 7

x cannot be 6

------------------------

Suppose x is 5

N + M + 3*5 = 19

N + M = 4

M must be 4 and N must be 0

The number is

05554 -cannot be

--------------------------------

Suppose x is 4

N + M + 3*4 = 19

N + M = 7

If M is 6, then N is 1

number is:

14446 -out of range

If M is 4, the N is 3 -also out of range

-----------------------------------------

Suppose x is 3

N + M + 3*3 = 19

N + M = 10

If M is 6, then N is 4

number is:

43336 

the number is in range

-----------------------------

If x is 2, then

 M + N + 6 = 19

M + N = 13

52228 works

62227 works

---------------------

If x is 1

M + N + 3*1 = 19 

M + N = 16

this makes number out of range

( N is 10 or greater )

---------------------------------------

The possible numbers are:

43336

52228

62227

-----------

get a 2nd opinion if needed























 

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