Pre-RMO 2014/1
A natural number is such that What is the largest prime factor of ?
This note is part of the set Pre-RMO 2014
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Pranshu Gaba
6 years, 8 months ago
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442<2014<(44+1)2. This satisfies the above inequality. Therefore, k=44. The largest prime factor of 44 is 11. 11 is the answer.
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.
11 is the answer as the valuebof k is 44. Prime factorization of 44= 2 \times 2 \times 11
11
<=> 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
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.
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
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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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
11
11 is the answer
442=1936<2014<452=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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Easy....11
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