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

Correct answer

12

Step-by-step solution

aligned & 166=b^2+6 b+6 & 56=5 b+6 & 8590=8 b^3+5 b^2+9 b & Now, (b^2+6 b+6 )(5 b+6)=8 b^3+5 b^2+9 b & 5 b^3+36 b^2+66 b+36=8 b^3+5 b^2+9 b & 3 b^3-31 b^2-57 b-36=0 & (b-12) (3 b^2+5 b+3 )=0 & b=12 aligned We have only one b which is 12 So, sum =12

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