Let aaa and ddd be two coprime integers. (The great common divisor between aaa and ddd is 111). Prove that: (∀n∈N∗):1a+1a+d+1a+2d+⋯+1a+nd∉Z(\forall n\in\mathbb{N}^*):\frac{1}{a}+\frac{1}{a+d}+\frac{1}{a+2d}+\cdots+\frac{1}{a+nd}\notin\mathbb{Z}(∀n∈N∗):a1+a+d1+a+2d1+⋯+a+nd1∈/Z.
Note by Mountassir Farid 7 years, 9 months ago
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Why is gcd(a,d)=1\gcd(a,d)=1gcd(a,d)=1? If this holds, then it also holds for a,da,da,d with gcd(a,d)>1\gcd(a,d)>1gcd(a,d)>1.
You could edit title as Sum of an Harmonic Progression....
What about a=1a=1a=1, d=1d=1d=1, and n=∞n=\inftyn=∞? Or is that not allowed? EDIT: would ∞∈Z\infty\in\mathbb{Z}∞∈Z?
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∞∉N ⟹ ∞∉Z. \infty \not\in \mathbb{N} \implies \infty \not\in \mathbb{Z}. ∞∈N⟹∞∈Z.
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Easy Math Editor
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.
When posting on Brilliant:
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> This is a quote
\(
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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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Why is gcd(a,d)=1? If this holds, then it also holds for a,d with gcd(a,d)>1.
You could edit title as Sum of an Harmonic Progression....
What about a=1, d=1, and n=∞? Or is that not allowed? EDIT: would ∞∈Z?
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∞∈N⟹∞∈Z.