Expert: Scott A Wilson - 11/7/2009

Question
QUESTION:  Problem: The sum of unused digits is 16

T E N T +
S I T E
__________
S L E E P

ANSWER: S has to be a 1, since there is 1 more digit in the answer and the sum of two distinct digits is no more than 17, making a 1 be carried.

If T were 9, then something wouldn't carry from from N+T=E, which would make I=9,
but T was already 9 in this case, so that wouldn't work.

If T=9, E+I must carry 1, so that makes L into a 0 since S is a 1.
This makes the starting numbers L=0, S=1, and T=8.

If E were 2, for E+I to carry one, I would have to be 9.  
That's one choice to look into, though most likely it won't work.

Since E + I = E, I must be 0 and nothing was carried or I must be 9 and one was carried.
Since we already have L as 0, I must be 9.

So far the ones we have determined are L=0, S=1, T=8, and I=9.
This gives 8EN8 + 198E = 10EEP.

If E=2, 8+E=10, so P=0, but 0 is taken.  No go.
If E=3, 8+E=11, so P=1, but 1 is taken.
If E=4, 8+E=12, so P=2.
Carry 1, so 1+N+8=14, so N=5.

Following the same idea and trying E to be other numbers gives me
L    0    0    0
S    1    1    1
P    4    2    4
E    6    4    5
N    7    5    6
T    8    8    8
I    9    9    9

That means there are 3 choices I found, one in each of the columns.


---------- FOLLOW-UP ----------

QUESTION: Scott I forgot to ask you this. where is " The sum of unused digits is 16" used here in your answer.

Which ones are unused digits anyway? I assumed T I N P because they were not used in the result??? I am probably wrong :(

Answer
The three sets of numbers I gave you were
A) 0,1,3,6,7,8,9
B) 0,1,2,4,5,8,9 and
C) 0,1,4,5,6,8,9 for L,S, P, E, N, T, and I.

Note that the numbers not used for set (A) are 2, 4, and 5, whose sum is 11.
Note that the numbers not used for set (C) are 2, 3, and 7, whose sum is 12.

The set that is wanted is set (B), for they are 3, 6, and 7, whose sum is 16.