# Digits A, B, C

For the various digits A, B, C is true: the square root of the BC is equal to the A and sum B+C is equal to A. Calculate A + 2B + 3C. (BC is a two-digit number, not a product).

Result

A+2B+3C =  28

#### Solution:

$0 \le A,B,C \le 9 \ \\ \sqrt{BC}=A \ \\ B+C = A \ \\ \ \\ 10B+C = A^2 \ \\ A = 9 \ \\ B = 8 \ \\ C = 1 \ \\ A+2B+3C= 9+2\cdot 8+3\cdot 1 = 28 \ \\$

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