# Understand the problem

##### Source of the problem

# 2018 AMC 10A/Problem 25

##### Topic

##### Difficulty Level

##### Suggested Book

# Start with hints

Comparing coefficients we have \( c= \frac {a^2} {9} \) and \( c-b= – \frac {a^2} {9} \Rightarrow b= \frac {2a^2} {9} \) . So , \( a+b+c = a + \frac {a^2}{3} \) . As \( a, b \ and \ c \) are nonzero digits \( \Rightarrow a + \frac {a^2}{3} \) is an integer \( \Rightarrow 3|a^2 \) \( \Rightarrow 3|a \) . Now to maximize \( a+b+c = a + \frac {a^2}{3} \) , put highest value of \( a \) i.e. \( 9 \) but it will give \( b= 18 \) which is not possible . [ as \( a, b \ and \ c \) are nonzero digits ] . Then putting \( a= 6 \) , \( max (a+b+c) =18 \) .

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