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Question

For xR, let [x] denote the greatest integer x, then the sum of the series [13]+[131100]+[132100]+....+[1399100] is :

A
135
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B
153
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C
131
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D
133
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Solution

The correct option is D 133
[13]+[131100]+[132100]+....+[1399100]
where xR,[x] denote the greatest integer x,
Since [13]=1 and
13x100=100+3x300
therefore,
For case 1-
100+3x<300
x<66.67[13x100]=1
For case 2-
300100+3x<6002003x<500367x<166
[13x100]=2
So the sum of the series equals
=(11167 times)+(22233 times)
672(33)=133

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