Determine the number of positive integers n that satisfy:

1/2 < n/n+1 < 99/101

I don't know how to solve this besides plugging in random numbers, which would take all day. Any other suggestions for a faster way to solve it?

Thank you!

The smallest number that satisfies

1/2 < (n/n+1) is 2/3
The largest number that satisfies
n/(n+1) < 99/101 is 49/50.
Here's proof of that:
49/50 = 0.980000
99/101 = 0.980198
50/51 = 0.98039

So all n/(n+1) numbers in the series
2/3, 3/4, 4/5 ... 49/50 satisfy the inequality.
There are therefore 48 numbers n that satisfy the condition.

maybe the largest is 99/100

To solve the inequality (1/2) < n/(n+1) < (99/101), we can follow these steps:

Step 1: Simplify the inequality.
First, cross-multiply the middle term n/(n+1) to get (n+1)n:

(1/2) < (n+1)n/(n+1) < (99/101)

Simplifying further:

(1/2) < n < (99/101)

Step 2: Convert the fractions to have a common denominator.
Multiply (1/2) and (99/101) by 101 to get:

101/202 < n < 99/1

Simplifying:

101/202 < n < 99

Step 3: Express n as a positive integer.
Since n is a positive integer, we can express the inequality as an integer inequality:

102/202 < n < 99

Step 4: Simplify the inequality further.
Divide both sides of the inequality by 2 to get:

51/101 < n < 99

Step 5: Count the number of positive integers in the given range.
To determine the number of positive integers n that satisfy the inequality, count the positive integers between 51/101 and 99.
The positive integers in this range are 52, 53, 54, ..., 98.

Therefore, the number of positive integers n that satisfy the inequality is 98 - 52 + 1 = 47.

So, there are 47 positive integers n that satisfy the given inequality.

To determine the number of positive integers that satisfy the given inequality, we can set up an equation and then solve it algebraically.

Let's rewrite the inequality:
1/2 < n/(n+1) < 99/101

To simplify the inequality, let's cross-multiply to eliminate the denominators:
(1/2)(n+1) < n < (99/101)(n+1)

Next, let's distribute:
1/2 * n + 1/2 < n < 99/101 * n + 99/101

Now, let's simplify the resulting inequality:
(1/2)*n + 1/2 < n < (99/101)*n + 99/101

We'll start with the lower bound:
(1/2)*n + 1/2 < n

Multiplying both sides of the inequality by 2 to eliminate the fraction:
n + 1 < 2n

Now, subtract n from both sides:
1 < n

So, the lower bound is n > 1.

Now, let's solve for the upper bound:
n < (99/101)*n + 99/101

Subtracting n from both sides of the inequality:
0 < (99/101)*n - n + 99/101

Simplifying the right-hand side by finding a common denominator:
0 < (99n - 101n + 99)/(101)

Combining like terms:
0 < -2n + 99/101

Adding 2n to both sides:
2n < 99/101

Finally, divide both sides by 2 to isolate n:
n < 99/202

So, the upper bound is n < 99/202.

Combining the lower bound and upper bound, we have:
1 < n < 99/202

To find the number of positive integers that satisfy this inequality, we need to count the integers within this range.

One approach is to subtract 1 from the upper bound and then count all the integers from 1 to the rounded-down upper bound. Let's do the calculations:

Upper bound (rounded down): 99/202 ≈ 0.4901 ≈ 0

Counting integers from 1 to 0: there are 0 integers.

Therefore, there are 0 positive integers that satisfy the given inequality.

Why did the scarecrow win an award? Because he was outstanding in his field!

To solve this inequality, we can start by simplifying the expression n/n+1.

Since n is a positive integer, we can rewrite the inequality as:

1/2 < 1 + 1/(n+1) < 99/101

Next, let's subtract 1 from all the terms to get rid of the fraction and simplify further:

1/2 - 1 < 1/(n+1) < 99/101 - 1

-1/2 < 1/(n+1) < -2/101

Now, we can invert the inequality by taking the reciprocal of each term:

-2/101 < (n+1) < -2/1/2

Simplifying further:

-2/101 < (n+1) < -4/2

Lastly, we subtract 1 from each term:

-2/101 - 1 < n < -4/2 - 1

-203/101 < n < -6/2

Notice that n must be a positive integer. However, there are no positive integers between -203/101 and -6/2.

Therefore, there are no positive integers n that satisfy the given inequality.