This question is odd

Let $f(n)=4n+1$ . Note that $f(n)$ is continuous and monotonically increasing (never decreases). Also, $f(0)=1$ and $f(100)=401$ , so every odd number from 1 to 401, inclusive, is achieved exactly once for $n$ in the interval $[0,100]$ . This is $\boxed{201}$ values.

As $n$ ranges over all real values in the interval $[0,100]$ , we need to find out for how many values of $n$ is $4n+1$ an odd integer. For $4n+1$ to become an integer, $4n$ needs to be an integer.If $n$ is an integer then definitely $4n$ is an integer since multiplication of integers follows the closure property. We know that, $2 \mid 4$ $\Rightarrow 2 \mid 4n$ Thus $4n$ is an even integer. But when we add $1$ to $4n$ it becomes an odd integer i.e; $2 \nmid 4n+1$ So, every integer of the form $4n+1$ is an odd integer, where $n \epsilon Z$ . In the interval $[0,100]$ , there are $101$ integers, thus there are $101$ odd integers of the form $4n+1$ in this interval. But, as $n$ ranges over all real values in the given interval, $4n$ can also be an integer when $n$ is of the form $\frac{4k+1}{4}$ , $\frac{4k+2}{4}$ or $\frac{4k+3}{4}$ for some $k \epsilon Z$ .
Now we need to find out of these three type of real numbers how many give us an even integer for $4n$ . Since if $4n$ is even then only $4n+1$ will be odd. $Case 1:$ $4 \times \frac{4k+1}{4}$ $=4k+1$ It is an odd integer since $2 \nmid 4k+1$ . $$ $Case 2:$ $4 \times \frac{4k+2}{4}$ $=4k+2$ $=2(2k+1)$ It is an even integer since $2 \mid 2(2k+1)$ . In this problem the value of $n$ i.e $\frac{4k+2}{4}$ should be in the inteval $[0,100]$ . There are 100 such real numbers ranging from $\frac{1}{2}$ to $\frac{199}{2}$ . Thus in this case we get $100$ odd integers of the form $4n+1$ in the given interval. $$ $Case 3:$ $4 \times \frac{4k+3}{4}$ $=4k+3$ $=2(2k+1)+1$ $=2k^{'}+1$ for some $k^{'} \epsilon Z$ . It is an odd integer since $2 \nmid 2k^{'}+1$ $=2 \nmid 4k+3$ Thus, there are no even integers in this case also. $$ Therefore there are a total of $101+100$ $=\boxed{201}$ Thus there are a total of $\boxed{201}$ odd integers of the form $4n+1$ when $n$ ranges over all real numbers in the interval $[0,100]$ .

4 n must be an even integer since even + odd = odd

4 n = 2 k
n = 0.5 k where k is an integer with interval [0,200]
There are 201 numbers from 0 to 200

Therefore, there are 201 values of n such that 4 n + 1 is an odd integer

n ranges between 0 and 100, so 4n+1 ranges between 1 and 401, and there is 201 odd integer in this interval!

We know that $4n + 1$ has to be an integer. Therefore $4n$ must be an integer. For that to happen, $n$ has to be at least a rational.

Being an odd integer, means that $4n + 1$ = $2k + 1$ , where $k \in \mathbb Z$ . Isolating $n$ gives us $n = \frac{k}{2}$ .

Now, because $0 \leq n \leq 100$ , $0 \leq k \leq 200$ . Counting every integer that is between 0 and 200, reveals 201 possible $k$ s.

Therefore, there are 201 possibles values of $n$ in $[0, 100]$ that holds the property.

An easy approach is checking from the end results (4n + 1) The range of results will be from 1 (4 * 0 + 1) and 401 (4 * 100 + 1)

We can then rephrase the problem to: How many odd integers are there between 1 and 401 inclusive. This can easily be calculated with Round down[(n+1)/2].

The number of odd numbers from 1 to 401 is Rounddown[ (401+1)/2 = 201

The values of n satisfying the condition are

0/4, 2/4, 4/4, 6/4, ......., 400/4

So there are 400/2 + 1 no. of values of n satisfying the condition.

So there are 201 possible values for n.

All odd integers are in the form $2k+1,$ for some integer $k.$ So, if $4n+1$ is an odd integer, there must exist some integer $k_0$ such that $4n+1=2k_0+1.$ Solving for $n,$ we get $n = \dfrac{k_0}{2}.$ Thus, the values of $n$ we want are exactly those that are one-half of an integer. In the given range, there are $\boxed{201}$ possible values of $n,$ specifically $\dfrac{0}{2}, \dfrac{1}{2}, \dfrac{2}{2}, \ldots, \dfrac{200}{2}.$ $\square$

If $4$ is multiplied by a number of the form $\frac {2k - 1}{4}$ , you'll get $2k - 1 + 1 = 2k$ , which is not odd. That means that the numerator must be even, and then the fraction will be expressed with $2$ as the denominator. $100 = \frac {200}{2}$ . That means that for the numerator, we can choose all the numbers from $0$ to $200$ . In this interval there are $201$ numbers. Hence, the answer is $\boxed {201}$ .

Let n be of the form n= $\frac{k}{4}$
The range of n is [0,100] so the range of k is [0,400]
This implies 4n+1=k+1
So for all even values of k, 4n+1 is odd.
So the possible values of k are 0,2,4,......,398,400.
Therefore there are 201 values of k possible.
Therefore there are 201 values of n possible.

If $4n+1$ is an integer, $4n$ is an integer too, so $n$ must be equal to a rational number $\frac{p}{q}$ where $q \mid 4$ ( $q$ divides $4$ ). If $q=4$ , $4n+1=p+1$ so $p$ must be even, therefore simplifying $p$ with $q$ , the fraction will have denominator $2$ . If the fraction has denominator $2$ , $4n+1=2p+1$ , so it will be odd for each $p$ integer; if $0 \le n \le 100$ and $n=\frac{p}{2}$ , $0 \le p \le 200$ , thus $p$ can assume $\boxed{201}$ integer values. In this are also included integer values of $n$ , when $p$ is even.

Note that n is a real number in the range [0,100]. 4n+1 odd <=> 4n even <=> 2n is an integer. So n can be 0, 1/2, 1, 3/2, ..., 100, these are 201 values.

4n+1, is an odd integer, means 4n is even,

which in turn means 4n/2 = 2n is any

integer. This further means n can be any

integer k, from 0 to 100 (101 possible

values), as well as k/2 for each k.

So k together with k/2 can have 201

possible values for n (note for k =0, k = k/2).

4n+1 is itselft odd with any +ve integer value. so, from 0 to 100 are solutions and 1/2,3/2, 5/2, ......, 199/2 satisfy. so there are all 201 solutions............

All the values of n such that n=x/4 where x is even satisfies the condition.

Since n is less than or equal to 100, x should be less than or equal to 400.

So the numbers 0/4, 2/4, 4/4, 6/4 ......400/4 satisfies the condition and the number of terms can be easily found using the concept of arithmetic progression

for 4n +1 to be an odd integer, 4n has to be an even integer.

For n=an integer, all 101 values from [0,100] will satisfy, moreover even the values of the form z+ 0.5, where z is an integer, will be 4z+2 when multiplied by four and thus will be even.Thus we have another 1 set of values, satisfying us. Thus the total no. of values will be 100+ 101=201

0 to 10.5 containing 22 numbers at an interval of 0.5, 11 to 20.5, 21 to 30.5 ... so on till 81 to 90.5, 91 to 100 so basically it is 22+20 *8 times and 19 for the last number line. is 201

If $4n+1$ must be an odd integer, then $4n$ must be an even integer.

For $n\in [0,100]$ , $4n$ ranges from $0$ to $400$ , inclusive. There are $201$ even numbers in the range $[0,400]$ , so there must be $\boxed{201}$ corresponding values of $n$ between $0$ and $100$ , inclusive.

First observe that 1 is an odd integer where n is chosen to be 0. Next observe that 401 is an odd integer where n is chosen to be 100. Every odd integer between 1 and 401 can be expressed in the form 4n+1 where n is a real value in (0, 100). Therefore this question is equivalent to counting the number of odd integers in [1, 401], which is simply just (401+1)/2 = 201.