An algebra problem by Rahil Sehgal

Algebra Level pending

A natural number $k$ is such that $k^2 < 2014 < (k+1)^2$ .

What is the largest prime factor of $k$ ?

The answer is 11.

1 solution

Achal Jain
Mar 10, 2017

This Question can be solved mentally.

it is given $(k+1)^{2} >2014> k^{2}$ . Now the first perfect square that comes to our mind( at least mine) is $2500$ which is square of $50$ . Now it's too much check, so check it for 45. We get $45^{2} =2025$ which is ''just'' greater than 2014 and just try it for $44^{2} \text{which is} 1936$ . Voila! the max value of $k$ is $44$ and $44=2 \times 2 \times 11$

$\therefore \text{ 11 is the answer}$

You can add it this way: Since, k^2 < 2014 < (k+1)^2, k< root(2014) < k+1 Here, we consider the closest integers of root(2014), i.e. 44, 45. [since root(2014) = 44.877...] Hence, the answer.

Shadman Zahid - 4 years, 3 months ago

An algebra problem by Rahil Sehgal

The answer is 11.

1 solution

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