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

## Check the Answer

But try the problem first…

Answer:$12$

PRMO-2018, Problem 4

Pre College Mathematics

## Try with Hints

First hint

Arrange the given number

Can you now finish the problem ……….

Second Hint

Arrange the equation \(166 \times 56 = 8590\) in base b where\(b\geq 10\)

Can you finish the problem……..

Final Step

Given condition is \(b \geq10\)

Now

\(166 = b^2 + 6b + 6\)

\(56 = 5b + 6\)

\(98590 = 8b^3 + 5b^2 + 9b\)

According to the problem,\((b^2 + 6b + 6) (5b + 6) = 8b^3 + 5b^2 + 9b\)

\(\Rightarrow 5b^3 + 36b^2 + 66b + 36 = 8b^3 + 5b^2 + 9b\)

\(\Rightarrow 3b^3 â€“ 31b^2 â€“ 57b â€“ 36 = 0\)

\(\Rightarrow (b â€“ 12) (3b^2 + 5b + 3) = 0\)

\(\Rightarrow b = 12\)

We have only one b which is 12

so the required sum is 12

## Other useful links

- https://www.youtube.com/watch?v=0yiR_y8YXc0
- https://www.cheenta.com/pattern-problem-amc-8-2002-problem-23/

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