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
3 years, 3 months ago
No vote yet
1 vote
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top NewestAn 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 \(a\).
Log in to reply
I'm an 7th grade student, so I just use original numbers :)) Sorry, can't understand.
Log in to reply
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
What is the i?
Log in to reply
\(i\) 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.
Log in to reply
Yes it is divisible!!!!!!!!!!!!!
Log in to reply