Cryptograms (Problem 2)

*Can be (and should be solved using number theory), but i tried to hard code it, since i am learning to code :) *

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#Initialise
r = 0
g = 0
b = 0

#Actual code
for r in range(9):
    for g in range(9):
        for b in range(9):
            if (300*r + 30*g + 3*b == 111*g):
                print(r,g,b)

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2

0 0 0
1 4 8

000 does not constitute distinct non-zero digits. Thus answer will be $GGG = \boxed{444}$

There are many ways to do this problem.

One way is to simply iterate through values of $B$ . There are only ten possibilities, and each forces a unique value of $G \equiv 3B \mod 10$ . One can then test to see whether this value of $G$ creates a contradiction in the tens' column.

For all of the following, I try each value of $B$ and let $G = 3B \mod 10$ .

Starting with $B = 1, G = 3$ ? This choice fails in the tens column because $3G = 9 \not\equiv 3 \mod 10$

Next, $B = 2, G = 6$ ? Fails because $3G = 18 \not\equiv 6 \mod 10$

And $B = 3, G = 9$ ? Fails because $3G = 27 \not\equiv 9 \mod 10$

Next, $B = 4, G = 6$ ? Note that we'll start to need to carry the excess from the ones' column to the sum in the tens column. In this case, it fails because $3G + 1 = 19 \not\equiv 6 \mod 10$

$B = 5, G = 5$ ? Fails because $B=G$ , which isn't allowed.

$B = 6, G = 8$ ? Fails because $3G + 1 = 25 \not\equiv 8 \mod 10$

$B = 7, G = 1$ ? Fails because $3G + 2 = 5 \not\equiv 1 \mod 10$

$B = 8, G = 4$ ? Then $3G + 2 = 24 \equiv 4 \mod 10$ . We then consider the hundreds column to find $R$ : $3R + 1 =4$ means $R = 1$ .

So $R = 1, G = 4, B = 8$ is a solution.

Just for sake of completeness, if $B = 9, 3B = 27$ , so $G = 7$ . But $3G + 2 = 23 \not\equiv 7 \mod 10$ .

So $(R,G,B) = (1,4,8)$ is the only solution.

Since $R\geq 1$ , therefore $G\geq 3$ .

$R=1,G=4,B=8, \overline {GGG}=\boxed {444}$

$\large{\begin{array}{ccccccc} & 1& 4&8 \\ & 1& 4&8\\ + & 1& 4&8\\ \hline & 4 & 4&4 \end{array}}$

Where $B = 8, G = 4, R = 1$