THREE + THREE + FOUR = ELEVEN
In the following addition sum the letters stand for numbers.
What are the numbers ?
THREE
THREE
FOUR
------
ELEVEN
Any clues ? we assume E=1, but the rest ?
Is your addition:
|
T |
H |
R |
E |
E |
|
|
T |
H |
R |
E |
E |
|
|
+ |
F |
O |
U |
R |
|
| - |
- |
- |
- |
- |
- |
|
| E |
L |
E |
V |
E |
N |
with E=1 and each of the other letters representing a digit form
2 to 9 or 0 and each digit being represented once only?
I think 0=R, 1=E, 2=N, 3=F, 4=H, 5=O, 6=V, 7=L, 8=T and 9=U works
(I think!)
Ian
Without much conviction, I also think that
78011
78011
5390
----
161412
also works.
Andre
Ian / Andre
Other than E=1 where there any other clues of logical deductions
you used to solve this problem ?
Regards
Jamie
I narrowed down the possibilities a little:
® To get a 1 under the U, U must be 8 (if 1 carried over from first
column) or 9 (otherwise).
R=9Þ N=1 ((no good))
R=8Þ U=8 ((no good))
Therefore R < 8 and nothing carried over hence U=9.
® I also looked at what could be 0. T can't be (Þ L is 0 or 1),
H can't be... I think I narrowed it down to U or R (and I haven't
checked U). I was wondering if R had to be 0.
I was fiddling around with little rules like this when I noticed my answer. No
doubt you can restrict further. Can you add some ideas Andre?
Ian
Ian, I approached it similar to yourself.
1) Depending on the carry (0 or 1) into the E,E,U column, U=9 or
8. If U=8 the carry into the E,E,U column must be 1. So R would
have to be 8 or 9, but R=8 duplicates U=8 and R=9 implies N=1.
Therefore U=9.
2) R can therefore take the values {0,2,3,4,5,6} (it can't be 7
else N=9). I looked at the consequence of assuming R=0 [if only
to eliminate it]. This implies N=2 therefore the remaining digits
are {3,4,5,6,7,8}.
So possible combinations of O and V are
(3,4),(4,5),(5,6),(6,7),(7,8). Similarly the possible
combinations of H,F are (3,5),(4,3),(8,5). This implies 15
combinations, which reduces to 7 combinations when you eliminate
repeated use of digits. For these 7, I computed what the implied
carry is into the T,T column, and checked whether the remaining
two digits (for T and L) worked from an arithmetic perspective.
Two combinations worked for (O,V),(H,F):
(5,6),(4,3)
(3,4),(8,5)
These give the solutions mentioned above.
If one were interested, one could explore other values for R. For
example, 46511+46511+8295=101317 appears to be a solution as
well.
Andre
Andre
I think the set of solutions is
46511+46511+8295=101317
74611+74611+2096=151318
78011+78011+5390=161412
84011+84011+3590=171612
85411+85411+0394=171216
It would have been a lot more satisfying had there just been one
solution ...
Andre
Thanks Andre (again).