Asked by Mr. Alexander
                Suppose $A$ and $B$ are digits for which the product of the two-digit numbers $\underline{2A}$ and $\underline{AB}$ is 1971. Find the sum of $A$ and $B$.
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Answers
                    Answered by
            Steve
            
    why bother wit LaTex? Just type it in:
Suppose A and B are digits for which the product of the two-digit numbers 2A and AB is 1971. Find A+B.
2A*AB = (20+A)(10A+B) = 1971
1971 = 3^3*73
So, its only two-digit factors are
27 and 73
A+B=10
    
Suppose A and B are digits for which the product of the two-digit numbers 2A and AB is 1971. Find A+B.
2A*AB = (20+A)(10A+B) = 1971
1971 = 3^3*73
So, its only two-digit factors are
27 and 73
A+B=10
                    Answered by
            nushi
            
    Suppose A and B are digits for which the product of the two-digit numbers 2A and AB is 1971. Find A+B.
2A*AB = (20+A)(10A+B) = 1971
1971 = 3^3*73
So, its only two-digit factors are
27 and 73
A+B=10
    
2A*AB = (20+A)(10A+B) = 1971
1971 = 3^3*73
So, its only two-digit factors are
27 and 73
A+B=10
                    Answer
            
    Suppose A and B are digits for which the product of the two-digit numbers 2A and AB is 1971. Find A+B.
2A*AB = (20+A)(10A+B) = 1971
1971 = 3^3*73
So, its only two-digit factors are
27 and 73
A+B=10
    
2A*AB = (20+A)(10A+B) = 1971
1971 = 3^3*73
So, its only two-digit factors are
27 and 73
A+B=10
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