Let E=12017+22017+32017+…+20162017 then E is divisible by
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answer is 2017.
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Detailed Solution
For 1≤x≤1013x2017+(2017−x)2017=x2017+2017C0(2017)2017 +2017C1(2017)2016(−x)+2017C2(2017)2015(−x)2+…+2017C2016(2017)(−x)2016+(−x)2017=(2017)mWhere m=2017C0(2017)2016+2017C1(2017)2015(−x) +…+2017C2016(−x)2016 is a positive integer.Now, write E as sum of 1013 groups as follow- F=12017+20162017+22017+20152017+…+10132017+10142017and note that each group is divisible by 2017