I have a new question for you all

Are you missing something? Median of what?

Suppose ten DISTINCT, POSITIVE integers have a median of x. ("DISTINCT integers" means that no two integers are the same.)

What is the smallest the average of those ten integers could be?

Explain your answer in complete sentences.

hint: Read every word in this problem carefully... especially the words "DISTINCT" and "POSITIVE"!

fixed yo' question

you're welcome.

After fixing your question, i think i can finally answer it.

suppose we have 10 different positive integers, a₁, a₂, ..., a₁₀, such that a₁ 2 3... 10 . the median of those numbers (x) is going to be (a ₅+a ₆)/2 (because a ₅ is the fifth number, and a ₆ is the sixth number, a ₅>=5 and a ₆>=6, so x can't be smaller than (6+5)/2=5.5. also x has to be equal to a positive number divided by 2, otherwise the question doesn't make any sense).

now, you want to minimize the average of those numbers.

first, we need to take a look at the numbers a₇, a₈, ....., a₁₀. those are the numbers that follow a₆. now, suppose a₆ is already known to us. this means that no matter what the numbers lower than a₆ are, their values do not depend on the values of a₇...a₁₀ because a₆ "seperates" them. so we want to minimize the average of our ten numbers, and because we know what the a₆ is, we can ignore the 5 smallest numbers, find the minimal possible sum of the 4 largest numbers, and THEN find the minimal value of the sum of the 5 smallest numbers.

so right now, our task is to minimize the sum of the 4 largest numbers. But you agree that if we can minimize EACH of the numbers a₇, ...,a₁₀ then the sum of the minimized numbers will be the minimal sum of the 4 numbers, right? (this is the part where you nod your head). so, we know that a₇ is an integer, and that it's larger than a₆, that is also an integer. this means that the minimal value of a₇ is a₆+1. we also know that a₈ is an integer and that it's larger than a₇, meaning that the minimal value for a₈ is a₇+1, and because the minimal value of a₇ is a₆+1, the minimal value of a₈ is a₆+2. we can do the exact same thing for a₉ and a₁₀ and conclude that the minimal value for a₇ is a₆+1, the minimal value for a₈ is a₆+2, the minimal valuer for a₉ is a₆+3 and the minimal value for a₁₀ is a₆+4. we know that it's possible for those 4 numbers to get those 4 values, meaning that the minimal sum of those 4 numbers is (a₆+1)+...+(a₆+4)=4*a₆+10.

Now let's look at the 4 lowest numbers. we can find the smallest possible value for their sum. so, by using the same trick i used for the sum of the 4 largest numbers we get that the smallest sum possible for the 4 lowest numbers is 1+2+3+4=10. we know that a₅>=5 so we know that those values dont affect a₅ or the numbers larger than a₅.

so, we know that no matter what, the smallest possible value for the sum of the 4 lowest numbers is 10. we know that the sum a₅+a₆ is x*2 (because x=(a₅+a₆)/2).  so we know what the minimal value for the sum of the 6 lowest integers is (10+2x) but we need to minimize the sum of the 4 largest integers. that sum is a₆*4+10, and this means that we need to minimize a₆. so we need to find the largest integer a₅ with the following conditions:

a) a₅>=5

b) 2x-a₅>a₅

c) there is no third condition

we can derive from (b) that a₅ 5 such that 5<=a ₅

(minimal sum of the 4 smallest integers)+(the sum of a₅+a₆)+(the minimal sum of the 4 largest integers)=

10+2x+(10+4*(2x-(largest a₅ such that 5<=a₅ and a₅​ is smaller than x))

20+10x-4*(largest a₅ such that 5<=a₅ and a₅ is smaller than x)

and that's the answer.