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# Problem based on Integer | PRMO-2018 | Problem 4

Try this beautiful problem from Algebra based on Integer

## Algebra based on Integer PRMO Problem 4

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.

• $24$
• $12$
• $34$

### Key Concepts

Algebra

Integer

base

Answer:$12$

PRMO-2018, Problem 4

Pre College Mathematics

## Try with Hints

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 $$b \geq10$$

Now

$$166 = b^2 + 6b + 6$$
$$56 = 5b + 6$$
$$98590 = 8b^3 + 5b^2 + 9b$$

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

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