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# Pre-RMO 2014/1

A natural number $$k$$ is such that $$k^2 < 2014 < (k + 1)^2.$$ What is the largest prime factor of $$k$$?

This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
2 years, 3 months ago

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$$44^2 < 2014 < (44+1)^2$$. This satisfies the above inequality. Therefore, $$k = 44$$. The largest prime factor of 44 is 11. 11 is the answer. · 2 years, 3 months ago

11 · 1 year, 4 months ago

11 · 1 year, 5 months ago

Easy....11 · 1 year, 5 months ago

11 · 1 year, 7 months ago

11 · 2 years, 1 month ago

$${44}^{2}$$=1936<2014<$${45}^{2}$$=2025,so if k <44 $$({k}+{1})^{2}$$<2014 therefore k=44 so the largest prime factor of k is 11. · 2 years, 2 months ago

11 is the answer · 2 years, 2 months ago

11 · 2 years, 3 months ago

k2<2o14 ..k<√2014 ...k<44.8 ..also ...2014<k2+2k+1 ...2014,(k+1)2 ...√2014<(k+1) ...44.8<(k+1) ...43.8<k ...k=44 · 2 years, 3 months ago

11 · 2 years, 3 months ago

11 · 2 years, 3 months ago

11 · 2 years, 3 months ago

Here k=44 satisfies the inequality.Now the prime factors of k are 2 and 11, of which 11 is the largest prime factor. · 2 years, 3 months ago

11 · 2 years, 3 months ago

11 · 2 years, 3 months ago

2014 lies between 44 and 45 square. But 45 square is 2025. Therefore we take 44. The prime factors of 44 are 2 and 11. The largest prime factor is 11 and smallest prime factor is 2. · 2 years, 3 months ago

k^2<2014. Max. value integer value of 'k' is 44 & it also satisfy the second equation i.e. (k+1)^2 >2014 . Hence largest value of k is 44 & its biggest prime factor is 11. So answer is 11. · 2 years, 3 months ago

<=> 0 < 2014 - k^2 < 2k +1 taking the right side of the inequallity and solving with the restrictions we get that the only integer is 44=4*11 therefore the answer is 11 · 2 years, 3 months ago

11 · 2 years, 3 months ago