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For a positive integer and nonzero digits , , and , let be the -digit integer each of whose digits is equal to ; let be the -digit integer each of whose digits is equal to , and let be the -digit (not -digit) integer each of whose digits is equal to . What is the greatest possible value of for which there are at least two values of such that ?

Number Theory

Medium

Mathematical Circles (Russian Experience)

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Use \( A_{n}= a(1+10+10^2….+10^{n-1}) = a \times \frac {10^{n -1}}{9} \) , similarly \( B_{n} = b \times \frac {10^{n -1}}{9} \) and \( C_{n} = c \times \frac {10^{2n -1}}{9} \) . Then proceed .

Using \( n > 0 \Rightarrow 10^n > 1 \) as \( n \in Z^+ \) , \( \ \) arrive at \( c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9} \) .

Observe that this expression \( c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9} \) is a linear combination of \( 10^n \) .
As per the question , \( c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9} \) is true for at least two distinct values of \( n \) .
\( \Rightarrow c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9} \) is an identity .
i.e. \( c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9} \) is true \( \forall n \in Z^+ \) . Then compare the coeffiecients .

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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