Suppose 1 1 + 2 1 + 3 1 + ⋯ + 2 0 1 0 1 = n m for some coprime positive integers m , n . Find the remainder when m is divided by 2 0 1 1 .
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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 n m = 2 0 1 1 4 0 2 0 . Obviously this is wrong, but can anyone explain why? Thanks.
It is true that n m ≥ 2 0 1 1 4 0 2 0 , but n m does not have to be equal to its minimum value.
And in fact, it isn't equal to its minimum value, since 1 = 2 = ⋯ = 2 0 1 0 .
Thanks for clearing my doubts..its torturing to be thinking of it nonstop. Thumbs up to you! :)
thanks for sharing this theorm
After this proof is the implication.
I've stumbled upon a weird problem: show that the sum of the multiplicative inverses up to (p-1)/2 is congruent to p 2 − 2 p modulo p.Any thoughts?
But p 1 is undefined ( m o d p ) , which means p 2 − 2 p is undefined ( m o d p ) , right? How is it possible that the sum is congruent to something undefined?
@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
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We only need to prove that 1 1 + 2 1 + ⋯ + 2 0 1 0 1 ≡ 0 ( m o d 2 0 1 1 )
This is because if n m ≡ 0 ( m o d 2 0 1 1 ) , 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 2 0 1 1 is prime, all the inverses are defined.
We'll prove that k 1 ≡ − 2 0 1 1 − k 1 ( m o d 2 0 1 1 ) , ∀ k ∈ [ 1 ; 2 0 1 0 ] ∪ N . This, since k , 2 0 1 1 − k are of different parity, will imply that the sum of the inverses ≡ 0 ( m o d 2 0 1 1 ) .
k k 1 ≡ − ( 2 0 1 1 − k ) k 1 ≡ 1 ( m o d 2 0 1 1 )
But, as you can see, this means that k 1 ≡ − 2 0 1 1 − k 1 ( m o d 2 0 1 1 ) - we have that k 1 is the inverse of − ( 2 0 1 1 − k ) , which is − 2 0 1 1 − k 1 .