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For xR, let [x] denotes the greatest integer ≤ x, then the value of 13+131100+132100+,,+1399100 is

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a
–100
b
– 123
c
–135
d
–153

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detailed solution

Correct option is C

For 0≤r≤66,0≤r100<23⇒ −23<−r100≤0⇒ −13−23<−13−r100≤−13∴ −13−r100=−1 for 0≤r≤66Also, for 67≤r≤100,67100≤r100≤1⇒ −1≤−r100≤−67100⇒ −13−1≤−13−r100≤−13−67100∴ −13−r100=−2 for 67≤r≤100Hence, ∑r=0100 −13−r100=67(−1)+2(−34)=−135


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