THREE + THREE + FOUR = ELEVEN

By Jamie Blake on Wednesday, July 24, 2002 - 08:16 pm:

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 ?

By Ian Short on Friday, July 26, 2002 - 02:11 pm:

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

By Andre Rzym on Friday, July 26, 2002 - 03:48 pm:

Without much conviction, I also think that

78011
78011
5390
----
161412

also works.

Andre

By Jamie Blake on Saturday, July 27, 2002 - 12:02 pm:

Ian / Andre

Other than E=1 where there any other clues of logical deductions you used to solve this problem ?

Regards

Jamie

By Ian Short on Monday, July 29, 2002 - 11:49 am:

I narrowed down the possibilities a little:

$\to$ To get a 1 under the $U$ , $U$ must be 8 (if 1 carried over from first column) or 9 (otherwise).

$R = 9 \Rightarrow N = 1$ ((no good))

$R = 8 \Rightarrow U = 8$ ((no good))

Therefore $R < 8$ and nothing carried over hence $U = 9$ .

$\to$ I also looked at what could be 0. $T$ can't be ( $\Rightarrow L$ 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

By Andre Rzym on Tuesday, July 30, 2002 - 10:19 am:

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

By Andre Rzym on Tuesday, July 30, 2002 - 03:25 pm:

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

By Ian Short on Wednesday, July 31, 2002 - 10:56 am:

Thanks Andre (again).