Answer:
8 5 2 9 1
+ 1 9 4 8 8
+ 3 5 3 8
1 0 8 3 1 7
As the hint tells you, T=1. You should be able to figure that out
pretty quickly anyway, because it was carried into the hundred-thousands
place. No matter how big E is, you can't carry anything more than 1
into the leftmost place.
In this solution, I'll use the shorthand that everything is
modulo 10 unless otherwise specified. That is, if I say that T+E+E=9,
you'll know I mean just that the one's digit is 9.
Also, I'll write the
carry in parentheses. For example, the tens column has H, E, and N
adding up to T. The carry into the tens column was 1, so I'll write
this as (1)+H+E+N=T.
Now that the ground rules are out of the way, let's
get to the puzzle...
E has to be a pretty big digit, at least 7, because when you add
1 to it (plus whatever got carried into the ten-thousands place) you get a
2-digit number. (But E can't be 9, because then T+E+E=Y, and so both
E and Y would be 9)
W, then, has to be 0, because (x)+E+T is at most
(2)+8+1=11. But it can't be 11, because then W would be 1, and the 1
is already taken. So (x)+E+1 is at most 10, so W is 0.
Y is 5 or 7, depending on whether E is 7 or 8, based on the one's
place. That is, if E=7 then T+E+E =1+7+7=5. If E=8 then T+E+E=1+8+8=7.
Here's what we know so far... If E=7 then Y=5
Case 1:
7 I G H 1
+ 1 H R 7 7
+ N I N 7
1 0 7 N 1 5
But this case doesn't work because even if 2 is carried into the
thousands column (2)+I+H+N can't be as high as 27 because I, H, and N can't be any more than 9, 8, and 6, in some
order. (2)+9+8+6=25, not enough to both leave a 7 in the thousands
place of the sum and generate the necessary carry of 2
into the ten-thousands place. So Case 1 is a non-starter.
If E=8 then Y=7:
Case 2:
8 I G H 1
+ 1 H R 8 8
+ N I N 8
1 0 8 N 1 7
In Case 2, 1 was carried into the tens place, so (1)+H+8+N=11 or
21. H and N can't be 1 or 0, so 11 is too low. (1)+H+8+N=21, so
H+N=12. 4+8 is ruled
out because 8 is already taken (E=8). 5+7 is ruled out because Y=7.
So H+N is 3+9, in some order. Here are the two orders:
Case 2a:
8 I G 3 1
+ 1 3 R 8 8
+ 9 I 9 8
1 0 8 9 1 7
All the digits are taken except for 2, 4, 5, and 6. From the
hundreds column we see that (2)+G+R+I=9, so G+R+I=7 (or 17 or 27). The
smallest possibility for G+R+I is 2+4+5=11, and the largest possibility is
4+5+6=15, so G+R+I=7 is impossible. Case 2a is ruled out.
Case 2b:
8 I G 9 1
+ 1 9 R 8 8
+ 3 I 3 8
1 0 8 3 1 7
In case 2b, 2, 4, 5, 6 are left for G, R, and I, and (2)+G+R+I=13, so they
are 2, 4, and 5. In the ten-thousands place, (1)+I+9+3=18, so I=5.
Case 2b:
8 5 G 9 1
+ 1 9 R 8 8
+ 3 5 3 8
1 0 8 3 1 7
The only digits left for R and G are 2 and 4. This is where the
second hint narrows the choice down to one possibility. Since R=2G,
we know R=4 and G=2, giving us the final answer.
