Prove that for any natural number n<2310 , n(2310−n) is not divisible by 2310.
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01.01.2015 09:23
clearly we see that. 2310=2.3.5.7.11 now it is clear that if 2310 divides n(2310−n) than it must divide n2 and hence each of the prime divisors of 2310 must divide n2 but than n2≥2.3.5.7.11=2310 a contradiction ! hence proved
01.01.2015 09:25
generalization of the problem let n be a natural number where n<k where k=p1p2p3...pm where p1,p2,.....,pm are distinct primes than prove that k does not divide n(k−n)
01.01.2015 09:44
Similar proof. By the way, it would be n2≥23102.
01.01.2015 09:46
yea thanks
24.05.2015 09:19
Proof of generalization either n\equiv0\pmod{p_i} or k-n\equiv-n\equiv0\pmod{p_i}, so either way n\equiv0\pmod{p_i}, so n\equiv0\pmod{p_1p_2\dots p_m} implying n\ge k.
23.09.2015 18:18
Let n be an integer which follows this property. Then since 2310 |2310n then it must also divide n^2 which in turn must also divide n becoz 2310=2×3×5×7×11. But n is not divisible by 2310 since n<2310,a contradiction. Hence, no solutions exist.Q.E.D......
19.11.2015 19:26
aditya21 wrote: clearly we see that. 2310 = 2.3.5.7.11 now it is clear that if 2310 divides n(2310-n) than it must divide n^2 and hence each of the prime divisors of 2310 must divide n^2 but than n^2 \geq 2.3.5.7.11 = 2310 a contradiction ! hence proved Hey please can someone please explain a bit more as I am not getting this one clearly..
19.11.2015 20:33
So ok we know 2310=2.3.5.7.11 so for n(2310)-n, 2310n is divisible by 2310 from question, \ n^2 is divisible by 2310^2 i.e its divisible by 2,3,5,7,11 which implies its divisible by 2310 hence n is greater than equal to 2310 but 2310 is greater than equal to n hene contradiction hope u understood