A discussion on an interesting number theory problem in RMO 2014

Problem :- Prove that there does not exist any integer n < 2310 such that n(2310-n) is a multiple of 2310.

I had written an alternative solution of this problem in the exam. Please go through the solution once and have your comments. Please inform about some improvements if necessary.

My solution :- We know that 2310 = 235711. So every prime factor in the prime factorization of 2310 occurs only once. So , suppose that n(2310-n) is divisible by 2310. Let p be a prime factor of 2310. So either p divides n or p divides 2310-n. Let us divide the problem into two cases :- Case 1:- p divides n :- If p divides n, then as p divides 2310, then p divides 2310-n. So n = pa and 2310-n = pb So n(2310-n) = (p^2)ab Hence (p^2) divides (2310-n)n Case 2 :- p divides 2310-n So, p must also divide n. Hence, in the same way, (p^2) again divides n(2310-n).

So we may conclude from both cases that if p divides 2310, then n(2310-n) is divisible by (p^2) So if 2310=abcd.....z where a......z are prime factors of 2310, then n(2310-n) = (a^2)(b^2)......(z^2)> (abc....z)^2 > (2310)^2 But we know that, n<2310 and 2310-n < 2310 So n(2310-n) < (2310)^2 So this leads to contradiction. Q.E.D.

Kindly comment on the solution.

#NumberTheory #RMO2014

Note by Siddharth Kumar
6 years, 5 months ago

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Comments

I think you don't have to divide the problem into two cases since the two cases are the same.

Poetri Sonya - 6 years, 5 months ago
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