+0

# help!

0
93
5

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

May 17, 2020

#1
+27998
+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

May 17, 2020
edited by ElectricPavlov  May 17, 2020
#3
0

That is correct, but wouldn't it help the student if you explained how you found the numbers?

Guest May 17, 2020
#4
+27998
0

See above....was working on explanation

ElectricPavlov  May 17, 2020
#5
0

Guest May 17, 2020
#2
0

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

.

May 17, 2020