Question about number theory - divisibility by $2^n$ ($n$ is large)

Question

Let $a,b,c,d,e$ five natural numbers ($a,b,c,d,e \in N$). We know that $2^{1386} | abcde$. (note that $a,b,c,d,e$ are not constant and we can choose them).

Find the largest $k \in N $ such that $2^k | a+b+c+d+e$.

note:the question didn't say $abcde = 2^{1386}$ but I think the answer is given by assuming this. So assume $abcde = 2^{1386}$

I know the answer is 280. But I don't know how to get to the answer. the first guess is 277 which $a=2^{277} , b=2^{277} , c = 2^{277} , d= 2^{277} , e= 2^{278}$ but it is not the right answer. I want to know how to get to 280. Please not just give an example and give a solution for finding that specific example or for finding $k = 280$.

Answer 1

For every $k\ge 278$ , we can choose $a=b=c=d=e=2^k$, doing the job.

Therefore, there is no largest $k$

Answer 2

We can build bigger powers of two from smaller ones, for example:

$$2^6 = 2^5+2^5 = 2^5+2^4+2^4 = 2^5+2^4+2^3+2^3 = 2^5+2^4+2^3+2^2+2^2$$

So in general this kind of 5-term partition gives

$$2^{k+4} = 2^{k+3} + 2^{k+2} + 2^{k+1} + 2^{k} + 2^{k}$$

It's clear that multiplying those $5$ terms of the addition gives $2^{5k+6}$. Do you see how to apply this to your problem?