How many positive integer values can x take that satisfy the inequality (x – 8) (x – 10) (x – 12)…….(x – 100) < 0?

Answer: 30

Explanation

Let us try out a few values to see if that gives us anything.

When x = 8, 10, 12, ….100 this goes to zero. So, these cannot be counted.

When x = 101, 102 or beyond, all the terms are positive, so the product will be positive.

So, straight-away we are down to numbers 1, 2, 3, …7 and then odd numbers from there to 99.

Let us substitute x =1,

All the individual terms are negative. There are totally 47 terms in this list (How? Figure that out). Product of 47 negative terms will be negative. So, x = 1 works. So, will x =2, 3, 4, 5, 6, and 7.

Remember, product of an odd number of negative terms is negative; product of even number of negative terms is positive. Now, this idea sets up the rest of the question.

When x = 9, there is one positive terms and 46 negative terms. So, the product will be positive.

When x = 11, there are two positive terms and 45 negative terms. So, the product will be negative.

When x = 13, there are three positive terms and 44 negative terms. So, the product will be positive.

and so on.

Essentially, alternate odd numbers need to be counted, starting from 11.

So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7…and then 11, 15, 19, 23, 27, 31,….. and so.

What will be the last term on this list?

99, because when x = 99, there are 46 positive terms and 1 negative term.

So, we need to figure out how many terms are there in the list 11, 15, 19,….99. These can be written as

4 * 2 + 3,

4 * 3 + 3,

4 * 4 + 3

4 * 5 + 3

4 * 6 + 3

…

4 * 24 + 3

A set of 23 terms. So, total number of values = 23 + 7 = 30. 30 positive integer values of x exist satisfying the condition.