Farina wrote:
S is the set of all prime numbers that satisfy the inequality
1|10−x|≤5x2What 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!
1|10−x|≤5x2The first thing we want to do is cross multiply since all the numbers here are positive:
x2≤5|10−x|Then square both sides to get rid of the absolute value:
x4≤25(10−x)2Now let's move
[25(10−x)2] to the left side of the equation:
x4−25(10−x)2≤0We have a difference of squares here, so let's factor this:
[x2−5(10−x)][x2+5(10−x)]≤0Distributing the 5's:
(x2−50+5x)(x2+50−5x)≤0And rearranging:
(x2+5x−50)(x2−5x+50)≤0We can factor the left expression, but we can't factor the right expression because its discriminant is less than 0:
(x+10)(x−5)(x2−5x+50)≤0So 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)≤02 satisfies this.
Let's try 3:
(13)(−2)(44)≤03 satisfies this.
What about 7?
(17)(2)(64)≤07 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:
2+3+55103Therefore the answer is 10