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.

## Comments

Sort by:

TopNewestAn 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

Log in to reply

– Adam Phúc Nguyễn · 2 years, 3 months ago

I'm an 7th grade student, so I just use original numbers :)) Sorry, can't understand.Log in to reply

– Prasun Biswas · 2 years, 3 months ago

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. :)Log in to reply

– Adam Phúc Nguyễn · 2 years, 3 months ago

What is the i?Log in to reply

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

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

Log in to reply

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

Log in to reply