If ⌊ x⌋ is the greatest integer less than or equal to x, then ∑_N=1^1024⌊log_2N⌋=

1986 AMC 12 Problem 25

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25.

\lfloor x\rfloor

is the greatest integer less than or equal to

x,

then

\sum_{N=1}^{1024}\lfloor\log_2N\rfloor=

8192

8204

9218

\lfloor\log_2(1024!)\rfloor

none of these

Group the integers

N

according to the interval

2^k\le N\lt2^{k+1}

There are

2^k

integers in the

k

th group; handle

N=1024

separately

For

0\le k\le9,

exactly

2^k

integers

N

satisfy

2^k\le N\lt2^{k+1},

and each contributes

k.

The final integer

1024=2^{10}

contributes

10.

Hence

\sum_{N=1}^{1024}\lfloor\log_2N\rfloor =\sum_{k=0}^{9}k2^k+10.

The finite geometric-sum identity gives

\sum_{k=0}^{9}k2^k=8194,

so the requested sum is

8204.

Thus the correct answer is B.