Farina wrote:
S is the set of all prime numbers that satisfy the inequality \(\frac{1}{|10-x|}\leq\frac{5}{x^2}\)
What is the value of the numerator of the fraction that is equal to the average (arithmetic mean) of S expressed in its most reduced terms?
Quant Sec 7
ID: Q02-01
This one is tough!
\(\frac{1}{|10-x|}\leq\frac{5}{x^2}\)
The first thing we want to do is cross multiply since all the numbers here are positive:
\(x^2≤5|10-x|\)
Then square both sides to get rid of the absolute value:
\(x^4≤25(10-x)^2\)
Now let's move \([25(10-x)^2]\) to the left side of the equation:
\(x^4 - 25(10-x)^2 ≤ 0\)
We have a difference of squares here, so let's factor this:
\([x^2 - 5(10-x)][x^2+5(10-x)]≤0\)
Distributing the 5's:
\((x^2 -50 +5x)(x^2 + 50 - 5x)≤0\)
And rearranging:
\((x^2 +5x - 50)(x^2-5x+50)≤0\)
We can factor the left expression, but we can't factor the right expression because its discriminant is less than 0:
\((x+10)(x-5)(x^2-5x+50)≤0\)
So now we need to find the prime numbers for the variable \(x\) that satisfies the inequality.
Note that 5 works, since (x-5) will go to 0. But are there others?
Let's try 2:
\((12)(-3)(44)≤0\)
2 satisfies this.
Let's try 3:
\((13)(-2)(44)≤0\)
3 satisfies this.
What about 7?
\((17)(2)(64)≤0\)
7 does not satisfy this.
From here, we can infer that as we increase \(x\), the inequality will continue to be unsatisfied.
So we have our three prime numbers of the set S: 2,3 and 5.
Now let's find the average of these numbers:
\(\frac{{2+3+5}}{5}\)
\(\frac{10}{3}\)
Therefore the answer is 10
03 Aug 2020, 00:12 
49% (01:22) correct
