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STAR + EAST = LIGHT
By Harriet Ter-Berg on January 6, 1999:

I found this question and its driving me mad who thought it up.

Its an easy question to write but how do you get the answer and is there a quicker way than just fiddling

The Star in the East gave great light each letter in the following sum stands for the digit 0,1,2,3,4,5,6,7,8.

can you work out the sum
.STAR
.EAST+ [The dots are for spacing, not decimals! - Ed]
-----
LIGHT.

I've worked out that R=0
L=1
S+E must equal 12,13,14 and 15 if its going to be a carry sum and I think it must be

Send help urgently

crazy of Norwich or Harriet T-B P60

By Rup on January 6, 1999:

If there is, I can't really see it... you're going to have to guess something. We know there's no carry in the units and there is a carry in the thousands; that only leaves four possible patterns of carry. Obviously (as you say) R=0 and L=1, and also none of A+S, S+E and A+T can equal 10 (else we'd need an R or another L on the bottom).

We also know that S+T+A+R+E+L+I+G+H = 0+1+2+...+8 = 36. That might help.

Case 1 -- no other carry
========================
so A+S=H, T+A=G and S+E=10+I. Substituting for A and S, get T+H+E=10+I+G.
In the sum of all letters, this gives S+A+2I+2G=25 or H+2(I+G)=25. Hence H is odd, but also the sum of A and S.

We only have three odd numbers to choose from, but it can't be 3 since that would have to be 1+2 and 1 is taken. H=5 means A and S are 2 and 3 but this won't leave a solution to S+E=10+I. Hence H=7 and so A=2, S=5. Now S+E=10+I, I>=3 so E=8 and I=3. Back to H+2(G+I)=25 means that G=6 -- which leaves T=4; yahoo!

Hence 012345678=RLAITSGHE; STAR=5420, EAST=8254 and LIGHT=13674.

I don't believe I can have guessed the right case first time -- are there other solutions for other carry patterns?

Case 2 -- carry in tens
=======================
A+S=10+H, T+A+1=G, S+E=10+I. Same attack again; T+H+E=I+G-1 and so H+2(I+G)=36. Now H is even; if H were 2, I+G would have to be 17 which is impossible -- as is 16 if H were 4. So if H is 6, I+G=15 so I is at least 7. However, S+E can then only be 11 not the 17 required. Hence H=8 -- but this is impossible, too: I+G=14 but the two highest left are 6 and 7. Contradiction.

Case 3 -- carry in hundreds
===========================
A+S=H, T+A=10+G, S+E=9+I. T+H+E=19+G+I, H+2(I+G)=16. Again H is even; the only remaining
solution to A+S=H even is 3+5=8 (didn't think of that last time :-) ) and so I+G=4. But that's clearly not possible with 2,4,6,7. Contradiction.

Case 4 -- carry in both tens and hundreds
=========================================
A+S=10+H, T+A=9+G, S+E=9+I. T+H+E=18+G+I, H+2(I+G)=17. For H odd, H can only be 7 as above; so I+G=5 and I and G are 2 and 3. However, 7=A+S so we need either 2+5 or 3+4 -- and we've already used both 2 and 3. Contradiction.

OK, so no other solutions. I guess that probably does amount to fiddling after all, but it was vaguely mathematical fiddling :-)

Rup.

By Max on January 11, 1999:

Hi, I've found the answer to this question.

As you predicted, SE has to be 12, 13, 14 or 15. R = 0. I started by choosing a random number for T. For T because it appeared the most. I decided that T could
not be very big, and not too small, so I randomly gave it a 4.

.S4A0
.EAS4+
-------
1IGH4

Then, I figured out that SE had to be something like 12, 13 or 15, since 4 was used. So making 1 and I as 2, 3, 5. I looked at the middle four letters: T A A S.
Since S had to be a part of 12, 13, 15, I decided to give it a 5.

.54A0
.EA54
-------
1IGH4

Now, for A, it had to be 7, 8. It couldn't be 2, 3 because it would not make a bigger number then 8. Returning to A, I gave it 2.

.5420
.E254
-------
1I674

Now, it fit perfectly. I had a 3 and a 8. I gave E the 8, so it would make a number more then 9. And so it made 13, where I put the 3 as I.

So.... S=5; T=4; A=2; R=0; E=8; L=1; I=3; G=6; and H=7.
There you go.

Max

By Alexander Maryanovsky on January 11, 1999:

There is no way to really find all the solutions without going through all the possibilities after you've found everything that you can "fish" out of the equation. The best way to find all the solutions is to run it through a computer.
There's a similar problem I know which has only one solution:

DONALD
+
GERALD
--------------
ROBERT

By The Editor:

For more alphanumeric puzzles like this, see the list in this discussion.