Cryptograms (Problem 3, Version 2)

$R_{max}(=98*3)=\boxed{2}94$ . Write the sum in another form: $30G+3B=11B+100R/:2$ $15G=4B+50R$ $5(3G-10R)=4B$ $3G-10R=\frac{4}{5}B$ Thus B=5. $3G=2(2+5R)$ $G=2\boxed{\frac{2+5R}{3}}$ Thus: $2+5R\;mod\;3=0$ $5R\;mod\;3=1$ $2R\;mod\;3=1$ $R=0;1$ are bad, thus $R=2$ . $G=2\frac{2+5\cdot2}{3}=2\cdot4=8$

We have $3(10G+B)=100R+11B\implies G=\dfrac{100R+8B}{30}, G\leq 9\implies 0<R\leq 2,0\leq B\leq 9,R\neq G, G\neq B, R\neq B$ .

The only such solution is $R=2,B=5,G=8$ , and $\overline {RBB}=\boxed {255}$ .

From the first look, $B$ can have two values either $0$ or $5$ . But if $B = 0$ , that means sum is a multiple of $100$ and also a multiple of $3$ because $GB$ is added $3$ times. So minimum possible sum would be $300$ . But $GB$ is a two digit number so $3$ times $GB$ is always less than $300$ . So no solution.

Taking $B = 5$ , we get

$3G + 1 = 10R + 5$

$\Rightarrow 3G = 10R + 4$

The only possible value of $G$ is $8$ with $R = 2$ .

So $RBB = 255$

$\large{\begin{array}{ccccccc} && 8&5 \\ && 8&5\\ +& & 8&5\\ \hline & 2& 5 & 5 \end{array}}$

where $B = 5, G = 8, R = 2$