To find number of numbers $\le100$ and divisible by $2$

Question

My textbook just writes as floor of $\dfrac{100}{2}$. How do I derive or understand this

How is this?

thanks

Answer 1

"floor of 100/2" means the smallest integer equals or larger than '100/2'. For example, "floor of 5.5" is 6, since 6 is the smallest integer equals or larger than 5.5

100/2=50, thus, "floor of 100/2" is 50.

And to this question, to find the number of numbers less than 100 and divisible by 2, we just need to find how many even numbers less than 100.

0 and 1 are smaller than 2, so no matter the lower bound is 0 or 1, the range is >=2, <=100.

from 2 to 100, we have 99 numbers, since the first one and the last one are divisible by 2, so 50 even number and 49 odd number.

Answer 2

The numbers divisible by $2$ are $2,4,6\ldots,100$
This forms an A.P. with first term $2$, common difference $2$ and last term $100$ $$a+(n-1)d=a_n\\2+(n-1)2=100\\n=50$$

Answer 3

If you want only to know how the answer is $\dfrac{100}2$, then from $1-100$ there are half of the numbers are even and half are odd.
Hence, the number of numbers divisible by $2$ is equal to the number of even numbers that is $$\frac{100}2$$

Answer 4

I'll try to give you a visual understanding. I do not have the time to draw stuff right now. So, I'm just gonna explain how to think about it. Maybe I'll update the answer later with some pictures.

Anyways, let's start. How do you visualize real numbers? There are many ways. But one of the most intuitive way is imagining them as points on the real line. Now, here we are dealing with integers. They can be visualized as points on the real line spaced by unit distance.

Where are the even numbers? They are just the points representing $0,\pm 2,\pm 4,\pm 6,\ldots$ spaced by a distance of $2$ units. It is easy to visualize the real line to be partitioned into semi-open intervals of length $2$ which are $\left( n, n+2 \right] \forall n\in \mathbb{2Z}$. Clearly, each of them has exactly one even number.

Back to our problem, we have the length $\left( 0 , 100 \right]$. The question of how many even numbers are there ultimate boils down to finding how many of such intervals fit in? Specifically, how many have their right hand side in it? (as that is the part containing the even number.) Since we start at $0$, we are starting at the left hand side of an interval. So, we need only find hoe many of such intervals belong to $\left(0,100\right]$ (completely, not partially, as the right hand side is all important). That's exactly the value of $\lfloor {100 \over 2} \rfloor$.