Categories
India Math Olympiad Math Olympiad PRMO

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

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

Subscribe to Cheenta at Youtube


Leave a Reply

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