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.
