Question 1133861
<br>
a+b+c = de<br>
The sum of three 1-digit integers is a 2-digit integer.  The tens digit of that 2-digit integer MIGHT be 2; but it is much more likely to be 1.  So let's assume d=1.<br>
de+d+e = fg<br>
If d=1, then this becomes 1e+1+e = fg, or 11+2*e = fg.  That means f has to be 2.  So now we know 11+2*e = 2g.<br>
fg+f+g = hi<br>
Since f=2, this becomes 2g+2+g = hi, or 22+2*g = hi.  That means h has to be 3.  So now we know 22+2*g = 3i.<br>
So digits 1, 2, and 3 are d, f, and h; so a, b, c, e, g, and i must be, in some order, digits 4 through 9.<br>
Now we now 11+2*e = 2g; and g can't be 1, 2, or 3.  That means e has to be 7, 8, or 9.<br>
And we know 22+2*g = 3i; and i can't be 1, 2, or 3.  That means g has to be 6, 7, 8, or 9.<br>
That's all the logical analysis I can see that we can do.  Now we just need to try sets of digits for a, b, and c for which the sum a+b+c is greater than 13 and less than 20; and see where that leads us.  For each set of digits we choose for a, b, and c, we stop looking at that case when a letter needs to be a digit that has already been used.<br><pre>
   a, b, c    e (units digit  g (units digit  i (units digit
                 of a+b+c)       of 11+2*e)      of 22+2*g)
  -----------------------------------------------------------
    456              5 X
    457              6               3 X
    458              7               5 X
    459              8               7               6  << YES; this should be the solution; but let's keep checking
    467              7 X
    468              8 X
    469              9 X
    478              9               9 X
    567              8               7 X
    568              9               9 X<br></pre>
There is indeed a single solution:  (a,b,c,d,e,f,g,h,i) = (9,5,4,1,8,2,7,3,6)<br>
a>b>c?  9>5>4?  YES
a+b+c = de?  9+5+4 = 18?  YES
de+d+e=fg?  18+1+8 = 27?  YES
fg+f+g=hi?  27+2+7 = 36?  YES