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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 $k = 280$.

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    I get why my example's k= 278 (not 279) because (1+1+1+1 + 2 = 6 which is divisible by 2) but I don't know how to use that. – 2017-01-17
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    Do you mean "choose integers $a,b,c,d,e$ so as to maximize $k$ subject to the constraints $2^{1386}|a.b.c.d.e$ and $2^k|a+b+c+d$ ? – 2017-01-17
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    @A.G. We don't want a,b,c,d,e. We want the k BUT It can be accomplished by choosing them too. But I don't want just five numbers. I want a way to get the k directly or a way to find that a,b,c,d,e. – 2017-01-17
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    I think the answer is wrong. Take for example $ a=b=c=d=e=2^{300}$, then $2^{1386}\mid 2^{1500} $ and $2^k\mid 5\cdot 2^{300} $ implies $ k=300$. – 2017-01-17
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    Something is missing here, even if you include the tacit assumption that $2^{1386}$ is the *largest* power of $2$ that divides $abcde$. Let $a=2^{1386}$, $b=c=d=a-1$, and $e=a+3$. Then $a+b+c+d+e=5a$, which is divisible by $2^k$ with $k=1386$. – 2017-01-17
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    Given the answer, it seems the question should possibly be "What is the largest $k$ such that for all naturals $n$ with $2^{1386}\mid n$, one can find naturals $a,b,c,d,e$ such that $2^k\mid a+b+c+d+e$ and $n=abcde$?" – 2017-01-17
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    @Xam - perhaps it should be the other way round, $abcde \mid 2^{1386}$ – 2017-01-17
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    @Joffan That would at least work - $2^{276}+2^{276}+2^{277}+2^{278}+2^{279}=2^{280}$ and $276+276+277+278+279=1386$ – 2017-01-17
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    @Joffan I think so, but I didn't ask the question. – 2017-01-17
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    @Xam note:the question didn't say $abcde = 2^{1386}$ but I think the answer (280) is given by assuming this. So assume $abcde = 2^{1386}$ – 2017-01-18
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    @HagenvonEitzen note:the question didn't say $abcde = 2^{1386}$ but I think the answer (280) is given by assuming this. So assume $abcde = 2^{1386}$ – 2017-01-18

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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?

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

Therefore, there is no largest $k$