Question 606856
Suppose the number is 9abc


We know that a = 2c, c > b. Also, 9+a+b+c = 20, so a+b+c = 11. We can assume c = 0, c = 1, c = 2, etc. and find all the possible triples (a,b,c):


c = 0 --> a = 0, b = 11
c = 1 --> a = 2, b = 8
c = 2 --> a = 4, b = 5
c = 3 --> a = 6, b = 2


The first three cases don't work, because b > 9, or b is greater than c. The third case does work, so (a,b,c) = (6,2,3) and the number is 9623.