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For each positive integer n, let σ(n) be the sum of its divisors (including 1 and n). Determine all positive integers n>1 such that there doesn't exist any positive integer m that satisfies n<m2≤σ(n).
We claim the only such numbers are the prime numbers and the square of prime numbers, excluding 3.
First, if p is a prime number, then σ(p)=p+1 and σ(p2)=p2+p+1. Then any such m satisfying p<m2≤σ(p) must satisfies m2=p+1, or p=m2−1. Except for p=3, this is a composite number (it has factors m−1,m+1). Any such m satisfying p2<m2≤σ(p) must satisfy m≥p+1 because m2>p2, but this leads to m2=p2+2p+1>p2+p+1=σ(p), contradiction.
Now, n=3 is a special case. Otherwise, if n is not a prime number or the square of one, then n has at least two additional distinct factors a,b which multiply to n. By AM-GM inequality, 2a+b≥ab or a+b≥2n; since a=b, equality is not achieved, so a+b>2n. Thus σ(n)≥1+a+b+n>n+2n+1=(n+1)2. But there must be a square number m2 between (n2,(n+1)2]; this satisfies n=n2<m2≤(n+1)2≤σ(n), contradiction.
@Ivan Koswara wow! nice solution ... how did yo think of tht ... Never seen such a problem .... Could yo provide links to such problems ..... Thanks :)
It's an Olympiad Problem, I don't know the link to this one in particular but if you search Math Olympiads online I'm sure you'll find many challenging problems. (If you're looking for even harder ones search IMO problems)
@IvaThank you very much, it's a very nice solution. I had been able to prove it doesn't work for primes except for 3 and for squares of primes, but i just couldn't prove that it works for every other number. Thanks! ;)
@Ivan Koswara Oh and if you like these problems, here's a harder one (I have literally made no progress on it):
Let n be a positive integer. We mean to paint every integer from 1 to n in blue or red. We say that the coloration is "balanced" if the following property is verified: if a and b are integers between 1 and n, painted in different colors, then, if a+b is also an integer between 1 and n, a+b is painted in blue, and if ab is also an integer between 1 and n, ab is painted in red.
Let N be the number of "balanced" colorations of the integers from 1 to n. Let P be the number of prime numbers p with the property that the largest integer smaller or equal to n/p is a power of p with a postive integer exponent. Prove that:
N=1+n+P.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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We claim the only such numbers are the prime numbers and the square of prime numbers, excluding 3.
First, if p is a prime number, then σ(p)=p+1 and σ(p2)=p2+p+1. Then any such m satisfying p<m2≤σ(p) must satisfies m2=p+1, or p=m2−1. Except for p=3, this is a composite number (it has factors m−1,m+1). Any such m satisfying p2<m2≤σ(p) must satisfy m≥p+1 because m2>p2, but this leads to m2=p2+2p+1>p2+p+1=σ(p), contradiction.
Now, n=3 is a special case. Otherwise, if n is not a prime number or the square of one, then n has at least two additional distinct factors a,b which multiply to n. By AM-GM inequality, 2a+b≥ab or a+b≥2n; since a=b, equality is not achieved, so a+b>2n. Thus σ(n)≥1+a+b+n>n+2n+1=(n+1)2. But there must be a square number m2 between (n2,(n+1)2]; this satisfies n=n2<m2≤(n+1)2≤σ(n), contradiction.
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@Ivan Koswara wow! nice solution ... how did yo think of tht ... Never seen such a problem .... Could yo provide links to such problems ..... Thanks :)
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It's an Olympiad Problem, I don't know the link to this one in particular but if you search Math Olympiads online I'm sure you'll find many challenging problems. (If you're looking for even harder ones search IMO problems)
@IvaThank you very much, it's a very nice solution. I had been able to prove it doesn't work for primes except for 3 and for squares of primes, but i just couldn't prove that it works for every other number. Thanks! ;)
@Ivan Koswara Oh and if you like these problems, here's a harder one (I have literally made no progress on it): Let n be a positive integer. We mean to paint every integer from 1 to n in blue or red. We say that the coloration is "balanced" if the following property is verified: if a and b are integers between 1 and n, painted in different colors, then, if a+b is also an integer between 1 and n, a+b is painted in blue, and if ab is also an integer between 1 and n, ab is painted in red. Let N be the number of "balanced" colorations of the integers from 1 to n. Let P be the number of prime numbers p with the property that the largest integer smaller or equal to n/p is a power of p with a postive integer exponent. Prove that: N=1+n+P.
@Aareyan Manzoor Can you help me solve these two other problems I've posted? Thank you.
wow i like math can you teach me?
(23)<(24)= (an>af) can you help me to solve that problems? thank you