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

Guest May 17, 2020

#1**+1 **

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

ElectricPavlov May 17, 2020

#2**0 **

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

_{.}

Guest May 17, 2020