Try this beautiful problem from Algebra based on Integer
The equation \(166 \times 56 = 8590\) is valid in some base \(b \geq10\) (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 \(b \geq10 \)satisfying the equation.
But try the problem first...
PRMO-2018, Problem 4
Pre College Mathematics
Arrange the given number
Can you now finish the problem ..........
Arrange the equation \(166 \times 56 = 8590\) in base b where\(b\geq 10\)
Can you finish the problem........
Given condition is \(b \geq10\)
\(166 = b^2 + 6b + 6\)
\(56 = 5b + 6\)
\(98590 = 8b^3 + 5b^2 + 9b\)
According to the problem,\((b^2 + 6b + 6) (5b + 6) = 8b^3 + 5b^2 + 9b\)
\(\Rightarrow 5b^3 + 36b^2 + 66b + 36 = 8b^3 + 5b^2 + 9b\)
\(\Rightarrow 3b^3 – 31b^2 – 57b – 36 = 0\)
\(\Rightarrow (b – 12) (3b^2 + 5b + 3) = 0\)
\(\Rightarrow b = 12\)
We have only one b which is 12
so the required sum is 12