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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

 May 17, 2020
 #1
avatar+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
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That is correct, but wouldn't it help the student if you explained how you found the numbers?

Guest May 17, 2020
 #4
avatar+27998 
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See above....was working on explanation cheeky

ElectricPavlov  May 17, 2020
 #5
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Thanks.  When I asked for an explanation, I hadn't seen that the other guest had already provided one.  Sorry . 

Guest May 17, 2020
 #2
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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 

.

 May 17, 2020

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