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I can think of a couple of ways

  1. Notice that unit digit of the first $8$ terms can be added in the last $8$ terms to make them $2017$. Now add the first $8$ terms without their unit digits (i.e. $2000*8$) and find a remainder on that.
  2. Sum of AP series = $4017*8$, now find the remainder

But both would take a lot of computation. Is there any more efficient way of doing it manually?

Nick P
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    why not work with negative remainders? – Shubham Johri Dec 24 '20 at 15:58
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    $-16-15-14-...-1=-1617/2=-817=-136\equiv1881\bmod2017$ – J. W. Tanner Dec 24 '20 at 16:00
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    BTW, the sum should be $4017*8$. – Shubham Johri Dec 24 '20 at 16:02
  • I wouldn't call $4017 \cdot 8$ a lot of computation. It is only four single digit multiplies, one of them including $0$, and a few carries. Using the negative remainders seems a bit easier to me, but not much. – Ross Millikan Dec 24 '20 at 16:04
  • @RossMillikan I understand your point-of-view, but it's all relative. It does seem like 'a lot of computation' to a grade 6 student who gets an average of 1 minute to do a question like this in a math competition. Also, a few seconds saved here can be used on another question. – Nick P Dec 24 '20 at 20:05

3 Answers3

4

It's $-16-15-14-...-1=-16\times17/2=-8\times17=-136\equiv1881\bmod2017$.

J. W. Tanner
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Welcome to MSE:

$2001 \equiv -16 \mod (2017)$

$2002 \equiv -15 \mod (2017)$

$2003 \equiv -14 \mod (2017)$

$\cdot$

$\cdot$

$\cdot$

$2016 \equiv -1 \mod (2017)$

Summing these relations we get:

$2001+2002+2002+2003+ . . . 2016 \equiv-\frac{16(1+16)}2=-136 \mod(2017)\equiv 1881 \mod(2017)$

J. W. Tanner
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sirous
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0

This is equal to: 2016(2017)(1/2)-(2000)(2001)(1/2) 2016(2017)(1/2)≡0(mod 2017)

2000(2001)(1/2)=(1000)(2001)≡(1000)(-16)≡-16000 (mod 2017) -16000 (mod 2017)≡-136(mod 2017)≡1881(mod 2017)

Asymptotic Joe
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