A cryptogram (part 4B)

Computer Science Level 4

$\begin{array} { l l l l l } & & A & B \\ & & C & D \\+ & E & F & G \\ \hline & H & I & J \\ \hline \\ \end{array}$

In the above equation, each of the letters $A, B, C, D, E, F, G, H, I$ and $J$ represents a different one of the digits 0, 1, 2, ..., 9.

Note that $A, C, E$ and $H$ cannot be 0.

How many possible answers for $\overline{HIJ}$ ?

The answer is 46.

1 solution

David Vreken
May 21, 2019

The following Python code finds $\boxed{46}$ possible answers for $HIJ$ .

# cryptogram4B
# Python 2.7.3

from itertools import permutations 
p = list(permutations([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], 10))
HIJlist = []
for k in range(0, len(p)):
    A = p[k][0]
    B = p[k][1]
    C = p[k][2]
    D = p[k][3]
    E = p[k][4]
    F = p[k][5]
    G = p[k][6]
    H = p[k][7]
    I = p[k][8]
    J = p[k][9]
    if A <> 0 and C <> 0 and E <> 0 and H <> 0:
        AB = 10 * A + B
        CD = 10 * C + D
        EFG = 100 * E + 10 * F + G
        HIJ = 100 * H + 10 * I + J
        if AB + CD + EFG == HIJ:
            if HIJ not in HIJlist:
                HIJlist.append(HIJ)
print "Possible answers for HIJ:", len(HIJlist)

But, isn't the length of p extensively large? i.e., 10 factorial.

Sathvik S - 1 year, 8 months ago

Yes, it is 10 factorial. This method has too many possibilities for a human to through, but takes most computers only a few minutes. That's why I solved it using a computer program.

David Vreken - 1 year, 8 months ago

A cryptogram (part 4B)

The answer is 46.

1 solution

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