Q.

Let E=12017+22017+32017+…+20162017 then E is divisible by

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a

2016

b

2017

c

4033

d

20162

answer is B.

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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)2016is a positive integer.Now, write E as sum of 1013 groups as follows:E=12017+20162017+22017+20152017+…+10132017+10142017and note that each group is divisible by 2017
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