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Tardigrade
Question
Mathematics
For x ∈ R, let [x] denotes the greatest integer ≤ x, then the value of [-(1/3)]+[-(1/3)-(1/100)]+[-(1/3)-(2/100)] +, ldots,+[-(1/3)-(99/100)] is
Q. For
x
∈
R
, let
[
x
]
denotes the greatest integer
≤
x
, then the value of
[
−
3
1
]
+
[
−
3
1
−
100
1
]
+
[
−
3
1
−
100
2
]
+
,
…
,
+
[
−
3
1
−
100
99
]
is
2008
220
Permutations and Combinations
Report Error
A
-100
B
-123
C
-135
D
-153
Solution:
For
0
≤
r
≤
66
,
0
≤
100
r
<
3
2
⇒
−
3
1
−
3
2
<
−
3
1
−
100
r
≤
−
3
1
∴
(
−
3
1
−
100
r
)
=
−
1
for
0
≤
r
≤
66
Also, for
67
≤
r
≤
100
,
100
67
≤
100
r
≤
1
⇒
−
1
≤
−
100
r
≤
−
100
67
⇒
−
3
1
−
1
≤
−
3
1
−
100
r
≤
−
3
1
−
100
67
∴
(
−
3
1
−
100
r
)
=
−
2
for
67
≤
r
≤
100
Hence,
r
=
0
∑
100
(
−
3
1
−
100
r
)
=
67
(
−
1
)
+
2
(
−
34
)
=
−
135