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$\begingroup$

I know that we first have to use AP to find the sum of first 2015 natural nos. Then we need to subtract all those factors of 2016. But there are too many. I need help in this part.

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    Here are divisors of 2016 : 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 28, 32, 36, 42, 48, 56, 63, 72, 84, 96, 112, 126, 144, 168, 224, 252, 288, 336, 504, 672, 1008. But you can actually find sum via http://mathschallenge.net/library/number/sum_of_divisors – 2017-01-04
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    $\phi(2016) = 576$ so this means that among the $2015$ fraction to inspect, only $576$ remain. – 2017-01-04
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    @Zubzub so what can i do with that....Like how should i proceed with that? – 2017-01-04

2 Answers 2

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Too long for a comment. I'm trying to explain the method used by @barak manos.

So we want : $$ \sum_{n=1|\ gcd(n,2016) = 1}^{2015} \frac{n}{2016} $$ It is clear that any time $n$ is a multiple of either $2$, $3$, or $7$ then $gcd(n,2016) > 1$ because the prime factorization of $2016 = 2^5 \cdot 3^2 \cdot 7$. So we need to remove all multiple of $2$, $3$ and $7$. But we have to be careful because if we remove once all multiples of $2$ and then all multiples of $3$ then we would have removed $2$ times the multiples of $6$ ! This is where the principle of inclusion/exclusion come into play :

  • Remove all multiples of $2$ and $3$ and $7$. This as for consequence to remove $2$ times the multiples of $6 = 2\cdot 3$, $14 = 2 \cdot 7$ and $21 = 3 \cdot 7$ and remove $3$ times the multiples of $42 = 2\cdot 3 \cdot 7$.
  • Add back one time the multiples $6,14,21$ that we removed one time too many. However this will add back $3$ times the multiples of $42$.
  • So overall we still need to remove the multiples of $42$ that we remove $3$ times but reintroduced $3$ times after.
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The prime factors of $2016$ are $2$, $3$ and $7$.

Hence use inclusion/exclusion principle:

  • Include $\sum\limits_{n=1}^{2016}\frac{n}{2016}$
  • Exclude $\sum\limits_{n=1}^{\frac{2016}{2}}\frac{2\cdot{n}}{2016}$
  • Exclude $\sum\limits_{n=1}^{\frac{2016}{3}}\frac{3\cdot{n}}{2016}$
  • Exclude $\sum\limits_{n=1}^{\frac{2016}{7}}\frac{7\cdot{n}}{2016}$
  • Include $\sum\limits_{n=1}^{\frac{2016}{2\cdot3}}\frac{2\cdot3\cdot{n}}{2016}$
  • Include $\sum\limits_{n=1}^{\frac{2016}{2\cdot7}}\frac{2\cdot7\cdot{n}}{2016}$
  • Include $\sum\limits_{n=1}^{\frac{2016}{3\cdot7}}\frac{3\cdot7\cdot{n}}{2016}$
  • Exclude $\sum\limits_{n=1}^{\frac{2016}{2\cdot3\cdot7}}\frac{2\cdot3\cdot7\cdot{n}}{2016}$
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    That's kinda "out of the sleeve" answer, totally unsure of it... – 2017-01-04
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    Well can u please explain why should we exclude some of the summations above – 2017-01-04
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    @user404047: Because we're double-counting them. That's the whole idea behind the principle of *inclusion/exclusion*. – 2017-01-04