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Suppose ten distinct, positive integers have a median of . ("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"!

sageatron2000 Apr 24, 2018

#2**0 **

Suppose ten __ DISTINCT__,

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 "

fixed yo' question

you're welcome.

Guest Apr 24, 2018

edited by
Guest
Apr 24, 2018

#3**0 **

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

suppose we have 10 different positive integers, a_{1}, a_{2}, ..., a_{10}, such that a_{1} 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 a_{7}, a_{8}, ....., a_{10}. those are the numbers that follow a_{6}. now, suppose a_{6} is already known to us. this means that no matter what the numbers lower than a_{6} are, their values do not depend on the values of a_{7}...a_{10} because a_{6} "seperates" them. so we want to minimize the average of our ten numbers, and because we know what the a_{6} 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 a_{7}, ...,a_{10 }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 a_{7} is an integer, and that it's larger than a_{6}, that is also an integer. this means that the minimal value of a_{7} is a_{6}+1. we also know that a_{8} is an integer and that it's larger than a_{7}, meaning that the minimal value for a_{8} is a_{7}+1, and because the minimal value of a_{7} is a_{6}+1, the minimal value of a_{8} is a_{6}+2. we can do the exact same thing for a_{9} and a_{10} and conclude that the minimal value for a_{7} is a_{6}+1, the minimal value for a_{8} is a_{6}+2, the minimal valuer for a_{9} is a_{6}+3 and the minimal value for a_{10} is a_{6}+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 (a_{6}+1)+...+(a_{6}+4)=4*a_{6}+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 a_{5}>=5 so we know that those values dont affect a_{5} or the numbers larger than a_{5}.

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 a_{5}+a_{6} is x*2 (because x=(a_{5}+a_{6})/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 a_{6}*4+10, and this means that we need to minimize a_{6}. so we need to find the largest integer a_{5} with the following conditions:

a) a_{5}>=5

b) 2x-a_{5}>a_{5}

c) there is no third condition

we can derive from (b) that a_{5} 5 such that 5<=a _{5}

(minimal sum of the 4 smallest integers)+(the sum of a_{5}+a_{6})+(the minimal sum of the 4 largest integers)=

10+2x+(10+4*(2x-(largest a_{5} such that 5<=a_{5} and a_{5} is smaller than x))

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

and that's the answer.

Guest Apr 24, 2018

edited by
Guest
Apr 24, 2018

edited by Guest Apr 24, 2018

edited by Guest Apr 24, 2018

edited by Guest Apr 24, 2018

edited by Guest Apr 24, 2018