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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

Check the Answer


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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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


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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