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Is \( 2 + 2^2 + 2^3 + 2^4 + ... + 2^{10}\) divisible by \(3\)?

It seems undetermined, because there were no threes, but let's try to pair it, extract it to find the threes.

\(A = (2 + 2^2) + (2^3 + 2^4)+...+ (2^9 + 2^{10})\)

\(A = 2(1+2) + 2^3(1+2) + ... + 2^9(1+2)\)

\(A = 2 \times 3 + 2^3 \times 3 + ... + 2^9 \times 3\)

\(A = (2 +2^3 + ... + 2^9) \times 3\)

Look! The series has x3! So A is divisible by 3.

Note by Adam Phúc Nguyễn
2 years, 4 months ago

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An easier way to prove that it is divisible by \(3\) is by using modular arithmetic. Let the given expression be \(Z\).

\[Z=\sum_{i=1}^{10} 2^i \equiv \sum_{i=1}^{10} (-1)^i\pmod{3}\\ \implies Z\equiv \left(\sum_{i=1}^5 (-1)^{2i}\right)+\left(\sum_{i=1}^{5} (-1)^{2i-1}\right)\pmod{3}\\ \implies Z\equiv \left(\sum_{i=1}^5 (1)\right)+\left(\sum_{i=1}^5 (-1)\right)\equiv \sum_{i=1}^5 (1-1)\pmod{3}\\ \implies Z\equiv \sum_{i=1}^5 (0)\pmod{3}\\ \implies Z\equiv 0\pmod{3}\\ \implies 3\mid Z\]

where \(a\mid b\) denotes that \(a\) divides \(b\), or in other words, \(b\) is divisible by \(a\). Prasun Biswas · 2 years, 3 months ago

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@Prasun Biswas I'm an 7th grade student, so I just use original numbers :)) Sorry, can't understand. Adam Phúc Nguyễn · 2 years, 3 months ago

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@Adam Phúc Nguyễn It's nothing much complicated. You can review the modular arithmetic wikis or ask me which part of it you can't understand?! I'd be more than glad to help you out. :) Prasun Biswas · 2 years, 3 months ago

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@Prasun Biswas What is the i? Adam Phúc Nguyễn · 2 years, 3 months ago

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@Adam Phúc Nguyễn \(i\) is a counter variable in the summation. Suppose, you have the sum \(S=2^1+2^2+2^3+\ldots+2^6\). You can write it as,

\[S=\sum_{i=1}^6 2^i\]

The initial and final values of \(i\) are given in lower and upper sides of the "sigma" symbol and the value of \(i\) gets incremented in each summation step. Prasun Biswas · 2 years, 3 months ago

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Yes it is divisible!!!!!!!!!!!!! Sudhir Aripirala · 2 years, 4 months ago

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