Solving Hard Single-Digit Multiplications Using an Internal Multiplication Table

At the bottom of this article is a table showing every possible value of an internal multiplication of the form:

..A...
x ______B_
..C...

particularly noting values where one or more of the letters are repeated.  This can be useful in solving certain single-digit multiplications.

OVERDUE       (2 words, 7851304629)
x ________E_
ARSENALS

(1) Here's a very easy example, where the possibilities are greatly restricted at the very start.  When the multiplier, internal digit, and internal product are identical, there are only four possibilities: 5, 6, 8, and 9.   In the example above, we can also see at once that E cannot equal 5 or 6, since  E x E = S at the far right.  E can also only equal 9 when the carry from the right equals 8, which would also force R to be 9.  Therefore E = 8, S = 4, and the carry from E x R is 4, so R equals either 5 or 6.  We also see that the carry from E x E is 6, and E x V = (S - 6) = 8, so V = 1 or 6 (of course V and R cannot both equal 6, so we have three cases for E/S/R/V).  Since E x O = (R - carry), we can calculate O, and only in the first case is there a valid value (8 x O cannot equal 9 or 5).  We thus have O = 3, and A = 2.  Now (8 x U) + 6 = L, and we can try the four unused values (0679) for U, to find that only U = 0 and L = 6 works.  There is no carry from E x U = L, so E x D = A, and D = 9.  The carry is 7, and (E x R) + 7 = 7, which is the last unused value, which equals N.

E S R V  (carry from ExV)  O A  U     L
8 4 5 1         1          3 2  0679  6xxx
8 4 5 6         6          x
8 4 6 1         1          x

(2) Now consider the problem below, inspired by Alice in Wonderland:

UGLIFI         (2 words, 0384196752)
x ______G_
CATION

As this is a short multiplication (the product is the same length as the multiplier), the possible values for U, G, and C are severely limited.  Either U is 1, or U is 2 and G is 3 or 4, or U is 3 or 4 and G is 2.   Considering the possible carries from the multiplication of G x G in the second leftmost column also gives us possible values for C.  If G = 2, there is no carry, and U cannot be 1 (since C = G x U, and C= G).  If U = 1, G cannot be larger than 6, otherwise the carry makes C greater than 9.  The possible sets of three values are shown in Table 1 below left, x indicating no possible unrepeated value):

U G C        G I              G A
1 2 x        2 09             2 45
1 3 4        3 0459           3 901
1 4 5        4 0369           4 6789
1 5 7        5 024579         5 56789
1 6 9        6 0123456789     6 678901
1 7 x
1 8 x        Table 2          Table 3
1 9 x
2 3 6
2 3 7
2 4 9
3 2 6
4 2 8

Table 1

For each value of G, we can extract from the full table at bottom all of the possible values of I (that is, the values where the product equals the internal digit of the multiplier).  These are marked in red on the table (and summarized in Table 2 above center).  Clearly we can omit all values where I = 0 or I = 1, or where I = 5 and G is odd, or where either = 6 and the other is even.  We list the others below, appending the value of N (from I x G = N) for each, and marking x where N repeats a value already assigned.  10 valid cases remain.  For each of these, we consider the carry to G x I = I, and list the unduplicated values of F which will produce the desired carry from F x G (these can be calculated on the fly or extracted from a Leading Digit Analysis table).  For example, on the first line, F x 3 produces a carry of 2 when F is at least 6, so we have unused values of 6 and 8.    On each line, we then multiply out F x G, adding the carry from I x G to the product to obtain the value of O.

U G C I   N  (carry   F  (carry      O
to GxG)      from IxG)
1 3 4 9   7     2     68     2       06
1 4 5 3   2     1     x
9   6     3     78     3       xx
1 5 7 2   0     2     4      1       x
4   0     4     89     2       2x
1 6 9 3   8     5     x
5   0     5     8      3       x
7   2     5     8      4       x
2 3 6 4   x
9   7     2     8      2       x
2 3 7 4   x
9   x
2 4 9 3   x
3 2 6 9   8     1     57     1       15
4 2 8 9   x

We then remove duplicate values and compress the table to the five valid lines, then add possible values for A (based on G x G = A), summarized above in Table 3.  Only four cases remain (one with two values of A).  With eight letters established in each case, we try the remaining values for L and check that L x G plus carry = T.

U G C I N F O A (carry from GxI)  L     T
1 3 4 9 7 6 0 x
1 3 4 9 7 8 6 0      2            25    xx
1 5 7 4 0 8 2 69     2            369   xxx
3 2 6 9 8 5 1 4      1            07    xx
3 2 6 9 8 7 5 4      1            01    1x

The correct values are on the last line, where L = 0 and T = 1, and the problem is solved. 