Sigma Floors

Algebra Level 4

$\displaystyle\sum _{ 1 }^{ 4096 }{ \left\lfloor \frac { \left\lfloor \frac { \left\lfloor \frac { \left\lfloor \frac { \left\lfloor \frac { n }{ 11 } \right\rfloor }{ 7 } \right\rfloor }{ 5 } \right\rfloor }{ 3 } \right\rfloor }{ 2 } \right\rfloor } =?$

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 1787.

1 solution

Chew-Seong Cheong
Mar 6, 2016

Let the sum be $S$ , then we have:

$\begin{aligned} S & = \sum_{n=1}^{4096} \lfloor \frac{\lfloor \frac{\lfloor \frac{\lfloor \frac{\lfloor \frac{n}{11} \rfloor}{7} \rfloor}{5} \rfloor}{3} \rfloor}{2} \rfloor \\ & = \sum_{n=1}^{4096} \left \lfloor \frac{n}{11\times 7 \times 5 \times 3 \times 2} \right \rfloor \\ & = \sum_{n=1}^{4096} \left \lfloor \frac{n}{2310} \right \rfloor \\ & = \sum_{n=2310}^{4069} 1 = 4096 - 2309 = \boxed{1787} \end{aligned}$

Nice!Learned a new property of floors today.

Rohit Ner - 5 years, 3 months ago

How can one prove this identity? Although I got the answer wrong ( silly mistake), I had the following approach:-

Let the term above 2 be A, the term above 3 be B, above 5 be C, above 7 be D.

To obtain a nonzero term, A must be greater than or equal to 2.

Let A=2, then B must be greater than or equal to 6.

Then, C would have to be greater or equal to 30.

Then D would have to be greater than or equal to 210.

Then N would be greater than equal to 210(11)=2310

Thus the least value of N is= 2310

Similarly for A=2, we obtain min(N)=4620

Thus All values of the term for n=2310 to 4096=1

Thus S= $\sum _{ n=2310 }^{ 4096 }{ 1 } =\quad 1\times 1787=\quad \boxed { 1787 }$

Aditya Dhawan - 5 years, 3 months ago

Work backward. Let $a = \lfloor \frac{b}{2} \rfloor$ . The minimum $b$ that satisfies the equation is $2a$ . Similarly, $b = \lfloor \frac{c}{3} \rfloor \Rightarrow c = 3b = 2 \times 3 a$ ; $c = \lfloor \frac{d}{5} \rfloor \Rightarrow d = 5c = 2 \times 3 \times 5 a$ ; $d = \lfloor \frac{e}{7} \rfloor \Rightarrow e = 7d = 2 \times 3 \times 5 \times a$ ; and $e = \lfloor \frac{n}{11} \rfloor \Rightarrow n = 11e = 2 \times 3 \times 5 \times 7 \times 11 a = 2310 a$ .

Therefore, for $a=1$ , the minimum $n = 2310$ , $a=0$ when $n < 2310$ and for $a=2$ , the minimum $n=4620$ . So we have:

$\begin{aligned} S & = \sum_{n=1}^{4096} \lfloor \frac{\lfloor \frac{\lfloor \frac{\lfloor \frac{\lfloor \frac{n}{11} \rfloor}{7} \rfloor}{5} \rfloor}{3} \rfloor}{2} \rfloor \\ & = \sum_{n=1}^{2309} \lfloor \frac{\lfloor \frac{\lfloor \frac{\lfloor \frac{\lfloor \frac{n}{11} \rfloor}{7} \rfloor}{5} \rfloor}{3} \rfloor}{2} \rfloor + \sum_{n=2310}^{4096} \lfloor \frac{\lfloor \frac{\lfloor \frac{\lfloor \frac{\lfloor \frac{n}{11} \rfloor}{7} \rfloor}{5} \rfloor}{3} \rfloor}{2} \rfloor \\ & = 2309(0) + (4096-2309)(1) = \boxed{1787} \end{aligned}$

Chew-Seong Cheong - 5 years, 3 months ago

That is a much better way of putting it. Thank you sir!

Aditya Dhawan - 5 years, 3 months ago

Sigma Floors

The answer is 1787.

1 solution

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