Topology

[cinelli]

Only in the field of alphametics could two Moebius strips combine to make a Klein bottle.

MOEBIUS
MOEBIUS+
-------
KBOTTLE

Cinelli

[ReformedCharacter]

Only in the field of alphametics could two Moebius strips combine to make a Klein bottle.

MOEBIUS
MOEBIUS+
-------
KBOTTLE

Cinelli

2051679
2051679+
- - - - - - -
4103358

RC

[ReformedCharacter]

Only in the field of alphametics could two Moebius strips combine to make a Klein bottle.

MOEBIUS
MOEBIUS+
-------
KBOTTLE

Cinelli

2051679
2051679+
- - - - - - -
4103358

RC

Sorry, that's wrong...

[GoSeigen]

Only in the field of alphametics could two Moebius strips combine to make a Klein bottle.

MOEBIUS
MOEBIUS+
-------
KBOTTLE

Cinelli

A mathematician named Klein
Thought the Möbius band was divine.
Said he: "If you glue
The edges of two,
You'll get a weird bottle like mine."

-Leo Moser

GS

[ReformedCharacter]

Only in the field of alphametics could two Moebius strips combine to make a Klein bottle.

MOEBIUS
MOEBIUS+
-------
KBOTTLE

Cinelli

I'll try again...

2948307
2948307+
- - - - - -
5896614

RC

[Gengulphus]

Only in the field of alphametics could two Moebius strips combine to make a Klein bottle.

MOEBIUS
MOEBIUS+
-------
KBOTTLE

Spoiler...

Look at the individual digit sums M+M, O+O, E+E, B+B, I+I, U+U and S+S. Each of them is an even number in the range 0 to 18.

Each of the result digits K, B, O, T, T, L, E is the units digit of the corresponding digit sum plus a possible carry in from the part of the addition to its right. That possible addition of a carry in can change a digit sum from 0 to 1, from 2 to 3, from 4 to 5, from 6 to 7, from 8 to 9, from 10 to 11, from 12 to 13, from 14 to 15, from 16 to 17, or from 18 to 19. Every one of those changes an even units digit to an odd one, and none of them alters whether the digit sum is >= 10, i.e. whether the column produces a carry out into the column to its left. I.e. each digit of KBOTTLE can be read off from the digit of MOEBIUS above it and the digit to its right according to the following table:

XY  where Z is determined by:  X       Y if Z<5  Y if Z>=5
XY                             ---------------------------
--                             0 or 5  0         1
Z                              1 or 6  2         3
                               2 or 7  4         5
                               3 or 8  6         7
                               4 or 9  8         9

The right hand column of the alphametic tells us that E must be even, i.e. one of 0, 2, 4, 6 or 8. Given whether B<5 or B>=5, we can use the third and second leftmost columns to determine O from the value of E, and then B from the value of O and whether E<5 or E>=5. There are ten possibilities for this, many of which don't work out:

E  B?   O  B  OK?
------------------------------------
0  <5   0  0  No - duplicated digits
0  >=5  1  2  No - B not in range
2  <5   4  8  No - B not in range
2  >=5  5  0  No - B not in range
4  <5   8  6  No - B not in range
4  >=5  9  8  Yes
6  <5   2  5  No - B not in range
6  >=5  3  7  Yes
8  <5   6  3  Yes
8  >=5  7  5  Yes

So B must be one of 3, 5, 7 or 8. B and I must be two different digits, but must produce the same value of T according to the first table above, so B and I must be one of the pairs for "X" in that table. That allows us to determine the value of I, as 8, 0, 2 or 3 respectively, and that in turn allows us to determine the value of T from the table, as 7, 0, 4 or 6 respectively. So the possibilities are now:

E  O  B  I  T  OK?
-------------------------------------
4  9  8  3  6  Yes
6  3  7  2  4  Yes
8  6  3  8  7  No - duplicated digits
8  7  5  0  0  No - duplicated digits

So we have just two possibilities left. At this point, take a look at the two corresponding possibilities for the whole alphametic:

M9483US
M9483US+
-------
K8966L4

M3672US
M3672US+
-------
K7344L6

M must be one of the digits 1, 2, 3 or 4, since 0 would be a leading zero and 5 or more would make the sum eight digits long. In the second possibility, 2, 3 and 4 have already been used, so M must be 1. But that makes K be 2, which has already been used, so we can eliminate the second possibility.

That just leaves the first possibility, in which 3 and 4 have already been used. So M must be 1 or 2, which makes K 3 or 5 respectively, and since 3 has already been used, it must be M=2 and K=5. Then S must be 2 or 7 and 2 has been used, so S=7, and it is then easy to check that the remaining digits 0 and 1 must be assigned to U and L as U=0, L=1, producing the unique solution:

2948307
2948307+
-------
5896614

Gengulphus

[ReformedCharacter]

Only in the field of alphametics could two Moebius strips combine to make a Klein bottle.

MOEBIUS
MOEBIUS+
-------
KBOTTLE

Spoiler...

Gengulphus

Which hut did you work in, at Bletchley?

RC

[cinelli]

I came upon this puzzle and solved it months ago. I made a note of the final sum but scant details of the method. So I have had to re-solve it – hence the delay in replying.

ReformedCharacter was the first with the correct solution and Gengulphus produced a complete method. Here is my solution and I leave it to the reader to decide how close my method is to that of Gengulphus.

Clearly E is even. Concentrate on EB + EB = TT with possible carry digits. A little thought shows that E and B differ by 5. Of the ten possibilities, BI = 50 can be dismissed as then T would be 0 too. There are nine possibilities for BI: 05, 16, 27, 38, 49, 61, 72, 83 and 94. The second and third columns of the sum link E, O and B. So consider the nine possibilities:

BI = 05, T = 1, O even = 0. Fails duplication.
BI = 16, T = 3, O even = 0, E = 5. Fails E must be even.
BI = 27, T = 5, O even = 6, E = 3. Fails E must be even.
BI = 38, T = 7, O even = 6, E = 3. Fails E must be even.
BI = 49, T = 9. Fails duplication.
BI = 61, T = 2, O odd = 3, E = 6. Fails duplication.
BI = 72, T = 4, O odd = 3, E = 6. A possibility.
BI = 83, T = 6, O odd = 9, E = 4. A possibility.
BI = 94, T = 8, O odd = 9. Fails duplication.

So we are left with just two possibilities for BI, 72 and 83:

M3672US
M3672US
-------
K7344L6

with digits not yet assigned: 0, 1, 5, 8, 9.

M9483US
M9483US
-------
K8966L4

with digits not yet assigned: 0, 1, 2, 5, 7.

The first arrangement can be dismissed because 2*M = K and not one of the available digits is double the other.

That leaves the second arrangement and it is clear that K = 5 and M = 2. S = 7, U = 0 and L = 1 quickly follow.

So the final sum is

2948307
2948307
-------
5896614

Cinelli