The equation $166 \times 56 = 8590$ is valid is some base $b \geq 10$ (that is $1, 6, 5, 8, 9, 0$ are digits in base $b$ in the above equation). Find the sum of all possible value of $b \geq 10$ satisfying the equation.

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Solution

$166 \times 56 = 8590$
$166 = x^{2} + 6 x + 6$ ……….. $(1)$
$56 = 5 x + 6$ ……… $(2)$
$8590 = 8 x^{3} + 5 x^{2} + 9 x + 0$ ………. $(3)$
$∴ (x^{2} + 6 x + 6) (5 x + 6) = 8 x^{3} + 5 x^{2} + 9 x$
$\Rightarrow 3 x^{2} - 31 x^{2} - 5 x - 36 = 0$
$(3 x^{2} + 5 x + 3) (x - 12) = 0$
$∴ 3 x^{2} + 5 x + 3 = 0$ (or) $x - 12 = 0$
$3 x^{2} + 5 x + 3 = 0$
$\to b^{2} - 4 a c = 25 - 4 \times 3 \times 3$
$= -$ ve roots are imaginary
$∴ x = 12$ only solution.

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