Proofs Relating to Integer Representation in Various Bases

furthermathishard

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Mar 19, 2020
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Hi ya'll, I'm currently really confused about a question that I met in a math exercise that's given by my teacher related to integers in other bases.
Posted below is the original question:
Screen Shot 2020-03-19 at 4.38.30 PM.png
I have already solved (probably incorrectly?) part (a), and am now attempting to write the proof for part (b)
I understand that I need to dissociate the base 3 into the factors that make it up, for instance
If N is even, then (anan-1 ... a1a0)3 = (2m)10
= 3^m x an + 3^m-1 x an-1...3a1+a0
= 3(3^m-1 x an + 3^m-2 x an-1...a1) + a0
Then I separated it into two cases:
1) if the sum of the terms in the bracket is odd
3 x even = even
even + odd/even = even
2) if the sum of the terms in the bracket is even
3 x odd = odd
odd + odd = even
odd + even = odd
so a0 is odd?

I have no idea how to continue from this point on, and this is only the proof from one direction ://
Please help if you know how to solve this!!
 
Before we move onto part b can we please see your work for part a?
 
I reiterate jomo's point, but ask as well did you try attacking part b by weak induction?
 
For part (b) I would begin by stating:

[MATH]N=\sum_{k=0}^n\left(a_k(2+1)^k\right)[/MATH]
And then using the binomial theorem:

[MATH]N=\sum_{k=0}^n\left(a_k\sum_{j=0}^k\left({k \choose j}2^{j}\right)\right)[/MATH]
[MATH]N=\sum_{k=0}^n\left(a_k\left(1+\sum_{j=1}^{k-1}\left({k \choose j}2^{j}\right)\right)\right)[/MATH]
Can you proceed?
 
To follow up, we may write:

[MATH]N=2\sum_{k=0}^{n}\left(a_k\sum_{j=1}^{k-1}\left({k \choose j}2^{j-1}\right)\right)+\sum_{k=0}^{n}\left(a_k\right)[/MATH]
The parity of the sum of an even number (the first sum as it is a multiple of 2) and the sum of the digits (the second sum) will only be even if the second sum is even.
 
Thank you guys so much!!
They've all inspired me in the process of solving this problem.
My teacher has graded my answers and I'll post it here! ?
WechatIMG42.jpeg
 
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