Categories

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

Try this beautiful problem from Algebra based on integer from PRMO 8, 2018. You may use sequential hints to solve the problem.

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

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.