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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
 #1
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Are you missing something? Median of what?

Guest Apr 24, 2018
 #2
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Suppose ten DISTINCTPOSITIVE 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.

Guest Apr 24, 2018
edited by Guest  Apr 24, 2018
 #3
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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‚Äč is smaller than x))

 

20+10x-4*(largest a5 such that 5<=aand a5 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

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