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0 votes
7.3k views
in Olympiad by (65.8k points)

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.

1 Answer

+2 votes
by (71.2k points)
edited by
 
Best answer

(166)a × (56)a = (8590)a

=> [a3 + 6a2 + 6] [5a + 6] = 8a3 + 9a2 + 5a

=> 3a3 – 31a2 – 57a – 36 = 0

=> (a – 12) (3a2 + 5a + 3) = 0

a = 12, only possible value of a.

by (10 points)
+1
But it's 8590 in the RHS, not 8950.

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