Please help with this hard problem.
Solve
\( \large{\begin{array}{ccccccc} & R& G&B \\ & R& G&B\\ + & R& G&B\\ \hline & B & B&B \end{array}} \)
R, G, B represent different, nonzero digits
b=5
g=8
r=1 is what I see
Whatever b is....when you add three of them you get b plus a carry digit.....only 5 will work for this 5 + 5 + 5 = 15 carry 1
then 1 + g +g + g must also end in a 5 8 is the only digit that works....with a carry of 2
then 2 + r + r + r must equal 5 also 1 is the onlt digit that works
Start with the column of B's
The only number that when multiplied by 3 ends with itself is 5
So B = 5
Then the column of G's
G times 3 has to end with 4 because you carry 1 from the units column.
The only number that when multipled by 3 ends with 4 is 8
So G = 8
The column of R's
You carry 2 from the tens column so R times 3 has to end with 3
The only number that will work is 1
So R = 1
Test it --
1 8 5
1 8 5
1 8 5
————
5 5 5
.