Remainders Are Fun 5

Algebra Level 3

Evaluate:

( i = 1 2017 p = 1 i 15 7 2 + 32 ( 8 p + 3 ) ( p + 20 ) ) m o d 1000 . \left( \sum_{i = 1}^{2017} \sum_{p = 1}^{i} \sqrt{157^2 + 32(8p + 3)(p + 20)}\right) \ \bmod 1000 \ .


For more problems like this, try answering this set .


The answer is 443.

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1 solution

Chew-Seong Cheong
Jun 28, 2017

S = i = 1 2017 p = 1 i 15 7 2 + 32 ( 8 p + 3 ) ( p + 20 ) = i = 1 2017 p = 1 i 24649 + 32 ( 8 p 2 + 163 p + 60 ) = i = 1 2017 p = 1 i 256 p 2 + 5216 p + 26569 = i = 1 2017 p = 1 i ( 16 p + 163 ) 2 = i = 1 2017 p = 1 i ( 16 p + 163 ) = i = 1 2017 ( 8 i ( i + 1 ) + 163 i ) = i = 1 2017 ( 8 i 2 + 171 i ) = 8 ( 2017 ) ( 2018 ) ( 4035 ) 6 + 171 ( 2017 ) ( 2018 ) 2 = 8 ( 2017 ) ( 1009 ) ( 1345 ) + 171 ( 2017 ) ( 1009 ) = ( 2017 ) ( 1009 ) ( 10760 + 171 ) = ( 2017 ) ( 1009 ) ( 10931 ) = ( 17 ) ( 9 ) ( 931 ) (mod 1000) = ( 17 ) ( 9 ) ( 69 ) (mod 1000) = 557 (mod 1000) = 443 (mod 1000) \begin{aligned} S & = \sum_{i=1}^{2017} \sum_{p=1}^i \sqrt{157^2+32(8p+3)(p+20)} \\ & = \sum_{i=1}^{2017} \sum_{p=1}^i \sqrt{24649 +32 (8p^2+163p+60)} \\ & = \sum_{i=1}^{2017} \sum_{p=1}^i \sqrt{256p^2+5216p+26569} \\ & = \sum_{i=1}^{2017} \sum_{p=1}^i \sqrt{\left(16p+163\right)^2} \\ & = \sum_{i=1}^{2017} \sum_{p=1}^i \left(16p+163\right) \\ & = \sum_{i=1}^{2017} \left(8i(i+1)+163i\right) \\ & = \sum_{i=1}^{2017} \left(8i^2+171i\right) \\ & = \frac {8(2017)(2018)(4035)}6 + \frac {171(2017)(2018)}2 \\ & = 8(2017)(1009)(1345) + 171(2017)(1009) \\ & = (2017)(1009)(10760+171) \\ & = (2017)(1009)(10931) \\ & = (17)(9)(931) \text{ (mod 1000)} \\ & = (17)(9)(-69) \text{ (mod 1000)} \\ & = -557 \text{ (mod 1000)} \\ & = \boxed{443} \text{ (mod 1000)} \end{aligned}

Nice Sir. :D

Christian Daang - 3 years, 11 months ago

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