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

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

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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= 2

211therefore largest prime factor of K is 11.

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11 is the answer as the valuebof k is 44. Prime factorization of 44= 2 \times 2 \times 11

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11

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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

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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.

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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.

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11

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11

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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.

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11

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11

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11

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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

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11

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11 is the answer

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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.

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11

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11

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Easy....11

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11

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