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You are here: Experts > Science > Math for Kids > Word Problems > Missing digits
Expert: Scott A Wilson - 11/4/2009
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.
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