Numerator Modulo 2011

Suppose 1 1 + 1 2 + 1 3 + + 1 2010 = m n \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{2010} = \dfrac{m}{n} for some coprime positive integers m , n . m, n. Find the remainder when m m is divided by 2011 2011 .


The answer is 0.

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2 solutions

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Mathh Mathh
Apr 11, 2015

We only need to prove that 1 1 + 1 2 + + 1 2010 0 ( m o d 2011 ) \frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{2010}\equiv 0\pmod {2011}

This is because if m n 0 ( m o d 2011 ) \frac{m}{n}\equiv 0\pmod {2011} , then n\cdot\frac{1}{n}\equiv 1\pmod {2011}\stackrel{\times m}\implies n\cdot \frac{m}{n}\equiv 0\equiv m\pmod {2011}

Since 2011 2011 is prime, all the inverses are defined.

We'll prove that 1 k 1 2011 k ( m o d 2011 ) , k [ 1 ; 2010 ] N \frac{1}{k}\equiv -\frac{1}{2011-k}\pmod {2011}, \forall k\in[1;2010]\cup \mathbb N . This, since k , 2011 k k, 2011-k are of different parity, will imply that the sum of the inverses 0 ( m o d 2011 ) \equiv 0\pmod {2011} .

k 1 k ( 2011 k ) 1 k 1 ( m o d 2011 ) k\frac{1}{k}\equiv -(2011-k)\frac{1}{k}\equiv 1\pmod {2011}

But, as you can see, this means that 1 k 1 2011 k ( m o d 2011 ) \frac{1}{k}\equiv -\frac{1}{2011-k}\pmod {2011} - we have that 1 k \frac{1}{k} is the inverse of ( 2011 k ) -(2011-k) , which is 1 2011 k -\frac{1}{2011-k} .

mathh mathh, we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments

Brilliant Mathematics Staff - 6 years, 2 months ago

We prove Wolstenholme's Theorem where if prime p > 3 and 1/1 + 1/2 + 1/3 + ... + 1/p-1 = m/n, then p | m.

We use multiplicative inverses to prove this out. Since the multiplicative inverses are unique modulo p, 1^(-1) + 2^(-1) + ... (p-1)^(-1) is congruent to 1 + 2 + ... + (p-1) modulo p and is congruent to (p-1)(p)/2. Since p > 3, p is odd and p-1 is even, hence, it is congruent to 0 modulo p implying that p | m.

I know this is a number theory question, but I tried to algebraically solve this question. In my approach, I used the AM-HM inequality and evaluate the equality case to give me m n = 4020 2011 \frac{m}{n}=\frac{4020}{2011} . Obviously this is wrong, but can anyone explain why? Thanks.

Shaun Loong - 6 years, 10 months ago

It is true that m n 4020 2011 \frac{m}{n}\ge \frac{4020}{2011} , but m n \frac{m}{n} does not have to be equal to its minimum value.

And in fact, it isn't equal to its minimum value, since 1 2 2010 1\neq 2\neq\cdots \neq 2010 .

mathh mathh - 6 years, 10 months ago

Thanks for clearing my doubts..its torturing to be thinking of it nonstop. Thumbs up to you! :)

Shaun Loong - 6 years, 10 months ago

thanks for sharing this theorm

math man - 6 years, 8 months ago

After this proof is the implication.

John Ashley Capellan - 6 years, 10 months ago

I've stumbled upon a weird problem: show that the sum of the multiplicative inverses up to (p-1)/2 is congruent to 2 2 p p \frac{2-2^p}{p} modulo p.Any thoughts?

Bogdan Simeonov - 6 years, 10 months ago

But 1 p \frac{1}{p} is undefined ( m o d p ) \pmod p , which means 2 2 p p \frac{2-2^p}{p} is undefined ( m o d p ) \pmod p , right? How is it possible that the sum is congruent to something undefined?

mathh mathh - 6 years, 10 months ago

@Mathh Mathh Well, the problem was given as the sum of the numbers, raised to the power of p-2, which is equal to their inverses, isn't it? here is the link

Bogdan Simeonov - 6 years, 10 months ago

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