$3+3+2+2+1=11$

Logic Level 4

$\large \begin{array} {c c c c c c } & T &H & R & E & E \\ & T & H & R & E & E \\ & & & T &W & O\\ & & & T &W &O\\ + & & & O &N & E \\ \hline E & L & E & V & E & N \\ \hline \end{array}$

In the above cryptarithm, find the value of $\overline{ELEVEN}$ .

$T$ , $H$ , $R$ , $E$ , $W$ , $O$ , $N$ , $L$ and $V$ represent distinct single digit non-negative integers, and $T,E$ and $O$ are non-zero.

The answer is 171219.

2 solutions

Sylvain Pelloquin
Aug 8, 2015

First, prove that E=1 : By absurd, if E=2, then T=9 and remainder=2 ; for this remainder to be 2, H must be 9 also... OK, then E=1.

Find triplets E/O/N : Let's call remainder of tens r1, of hundreds r2 and so on... Units column : 3 + 2×O = r1×10 + N Tens column : 2 + 2×W + N + r1 = r2×10 + 1 Eventually, O can't be > 3 because N is odd and if O>3, r1=1 and the tens sum becomes even (which is forbidden). You would say that O could be 9 so that r1=2 but in that case N becomes 1 (=E so forbidden) So O is 2 or 3. If O=2, then N=7 and W=6. If O=3, then N=9 and W=0 or 5

Pfew ! That was the easy part ! Next, you'll have to work on the H. You'll see that r3 can be only 1 or 3 (so that 2×H + r3=r4×10 + 1) If r3=1 : H=0 or 5 ; if r3=3 : H=4 or 9 For each of this 4 values of H, you'll have to find the possible values of T (secondarily L) which is a twelve or so :-) between 5 and 9...

Next you quietly try combination of triplets O/N/W (3 possibilities) with H/T/L (12). It won't make 36 possibilities because some triplets are incompatibles (for example O/N/W=3/9/0 and H/T/L=0/7/4). You'll have ten possibilities (or so...) to test for the hundreds column...

That was my pen-in-the-hand solution ! Feel free to comment...

Jesse Nieminen
Jul 1, 2019

from itertools import permutations

for T, H, R, E, W, O, N, L, V in permutations("0123456789", 9):
    if "0" in (T, E, O):
        continue
    if 2 * int(T + H + R + E + E) + 2 * int(T + W + O) + int(O + N + E) == int(E + L + E + V + E + N):
        print(E + L + E + V + E + N)

I have a question though; I am sorry this is so awkward in the future but if we had a problem like this:

-Then how would we solve it using permutations, because when I tried to just convert the letters to integers, it wouldn't come up with any result

Anonymous1 Assassin - 4 months, 3 weeks ago

3 + 3 + 2 + 2 + 1 = 11 3+3+2+2+1=11 3 + 3 + 2 + 2 + 1 = 1 1

The answer is 171219.

2 solutions

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$3+3+2+2+1=11$