$(x – 100) (x – 102) (x – 104)\cdots(x – 200) > 0$ for how many natural $x<250$?

Question

For how many natural numbers $x$ less than $250$ is the inequality: $$(x – 100) (x – 102) (x – 104) (x – 106)\cdots(x – 200) > 0$$ satisfied?

I tried to plot general points to get an idea of the problem, and could easily see the first positive point would be beyond x > 200, and subsequent sign changes will happen along all the points coming backwards. How do we compute all those ranges without actually manually plotting all 200 points on a number line, and seeing between what values of x the sign of inequality is positive?

Answer 1

There are $51$ zeros of the polynomial, leaving $52$ continuous segments where the polynomial is either always positive or always negative. Hence there are $26$ positive and negative segments each; the rightmost such segment must be positive since the leading coefficient is positive, and contains $249-201+1=49$ numbers. The other $25$ positive segments each contain a single integer each, as is easy to see; the polynomial's degree is odd, so the large leftmost segment is negative. This gives the final answer as $49+25=74$.

Answer 2

For any integer $x$ the left hand side is the product of $51$ integers. If $x>200$ then they are all positive, so the product is positive. If $x<100$ then they are all negative, and so the product is negative. For all even numbers inbetween, one of the factors is zero, so the product is zero. Can you see what happens at the odd numbers inbetween?

Answer 3

Well, basically that product is positive if no factor is $0$ (so $x<100$ or $x>200$ or $x=101,103,\ldots,199$) and an even number of factors are negative. Note that if $x>200$ all factors are positive, if $x<100$ all $51$ factors are negative, so the whole thing is negative. For the other $x$ the sign alternates, so for $101$ we then have positive, for $103$ we have negative and so on. So this gives us $25$ solutions. Adding the $49$ solutions with $200<x<250$ we get $74$ solutions.

Answer 4

Ok, let's see. Your product can be written as follows:

$\prod\limits_{k=0}^{50}(x-100-2k)>0$,

so you got an odd number of factors, a total of 51 products.

As you said before, for all $200<x<250$ the inequality is solved and you already get 49 solutions. If x<100 then you'll have an odd number of negative factors, so there will be a negative result which is not a solution of the inequality. Now we got two cases for $100\le{x}\le200$:

• x is even (e.g x=148), so your product will be $0$ and the inequality is unsolved;
• x is odd such that $x=101+2n$ for $n=0,...,49$ . If $n=0$ then $x=101$ and the first factor of the product is negative while all the others are positive, so the product will be negative. If $n=1$ then $x=103$ and the first two factors are negative while all the others are positive, so the product is positive. And so on...

You can conclude that for all odds $n$ between $0$ and $49\;$ $x=101+2n$ is a solution of the inequality, so in this case you got a total of 25 solutions. In conclusion:

The number of solutions of the product is given by 49+25=74 solutions.