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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, 9 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. Sharky Kesa · 2 years, 9 months ago

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To find Largest prime factor of any number K , you need to do its prime factorisation. for e.g.

prime factorisation of 10 is 10=5*2

so the largest prime factor is 10

But the question you asked is something like K^2<2014<(K+1)^2 if we take K^2<2014

we get k<+-44.88….. so largest value of K is 44

we take largest value so we can get largest prime factor. prime factorisation of 44 is -

44= 2211

therefore largest prime factor of K is 11. Ankit Kumar · 1 month ago

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11 Anunoy Chakraborty · 1 year, 10 months ago

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11 Gaurav Singh · 1 year, 11 months ago

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Easy....11 Akshat Sharda · 1 year, 11 months ago

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11 Mridul Jain · 2 years, 1 month ago

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11 Asish Sharma · 2 years, 8 months ago

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\({44}^{2}\)=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. Kristian Vasilev · 2 years, 8 months ago

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11 is the answer Vaibhav Sharma · 2 years, 9 months ago

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11 Vikram Bishla · 2 years, 9 months ago

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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 Wafaa Abd El Aziz · 2 years, 9 months ago

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11 Vishal Rai · 2 years, 9 months ago

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11 Zahra Y · 2 years, 9 months ago

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11 Palash Som · 2 years, 9 months ago

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Here k=44 satisfies the inequality.Now the prime factors of k are 2 and 11, of which 11 is the largest prime factor. Vivek Rao · 2 years, 9 months ago

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11 Shudipta _Cuet12 · 2 years, 9 months ago

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11 Subhajit Ghosh · 2 years, 9 months ago

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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. Harish Krishnan · 2 years, 9 months ago

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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. Rahul Verma · 2 years, 9 months ago

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<=> 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 Carlos David Nexans · 2 years, 9 months ago

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11 Pooja Deshmukh · 2 years, 9 months ago

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11 is the answer as the valuebof k is 44. Prime factorization of 44= 2 \times 2 \times 11 Pulkit Kapoor · 2 years, 9 months ago

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