From AIME 1994

(AIME 1994) Find the positive integer n for which
[log₂ 1] + [log₂ 2] + [log₂ 3] +…+ [log₂ n] = 1994
where [x] denotes the greatest integer less than or equal to x (for example [Π] = 3)

Solution:

Observe that:
[log₂ 1] = 0(2⁰)
[log₂ 2] + [log₂ 3] = 1(2¹)
[log₂ 4] + [log₂ 5] + [log₂ 6] + [log₂ 7] = 2(2²)
[log₂ 8] + [log₂ 9] +…+ [log₂ 15] = 3(2³)
[log₂ 16] + [log₂ 17] +…+ [log₂ 31] = 4(2⁴) ……

** [log₂ c] + [log₂ (c+1)] +…+ [log₂ (2c-1)] = b(2^b); where 2^b = c, for positive integers b and c.

So,
0(2⁰) + 1(2¹) + 2(2²) + 3(2³) + 4(2⁴) + 5(2⁵) + 6(2⁶) + 7(2⁷) + 456 = 1994

**7(2⁷) is the sum of [log₂ 128] + [log₂ 129] + [log₂ 130] +…+ [log₂ 255]. Hence, 456 = [log₂ 256] + [log₂ 257] + [log₂ 258] +…+ [log₂ n].

Since [log₂ 256]=[log₂ 257]=[log₂ 258]=…=[log₂ n] = 8
So, 456/8 = 57; Then n = 256 + (57-1).

Therefore, n = 312