 How Cheenta works to ensure student success?
Explore the Back-Story

# 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

## Subscribe to Cheenta at Youtube

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.

### Knowledge Partner  