Question

A set contains all possible five digit positive integers that can be made using each of the digits 1,3,5,7, and 9 once. Ex of possible integers: 13579; 35791;etc.

What is the median of that set? (not the example, but all of it)

Answer 1

Pleeeease help, someone????

Answer 2

I would start by subtracting the smallest possible number from the largest and then dividing by 2.

But to premise that, is the smallest number 11111 or is it 13579?

Answer 3

can only use each digit once

Answer 4

13579 would be first

97531 would be last

Answer 5

wouldn't the number of possible numbers
using 5 numbers 5*4*3*2*1 ?

Answer 6

Drat. My brilliant plan has resulted in 41976

I think I may have bitten off more than I can chew here... :-(

Answer 7

I know, this problem is hard and I'm trying my best to solve it and ask others too:)

Answer 8

well we know this for sure

there are 5*4*3*2*1 = 120 different ways to arrange the 5 numbers

Answer 9

the median will be in between locations 60 and 61 (the smallest number 13579 is at location 1 and the largest number 97531 is at location 120)

Answer 10

actually it will be the average of the values at locations 60 and 61

Answer 11

Thank you:)

Answer 12

the first 24 values start with a 1 (since 4! = 4*3*2*1 = 24)

the next 24 values start with a 3

so that's 24+24 = 48 values so far that are below the median

Answer 13

so the median must start with a 5 because 48+24 = 72 telling us that we've overshot locations 60 and 61

Answer 14

That makes sense now:) Thanks!

Answer 15

yw

Answer 16

@jim_thompson5910 You have a knack for bringing highly illuminating forehead slappers to the table! :-)

This is really interesting, and after studying your claims, I'll try reiterating them (verbosely) in a way I hope will help others to understand. I only hope this won't make things MORE confusing to anyone.

As you said, There are 5*4*3*2 = 120 ways of arranging the digits: 1,3,5,7,9

If you divide these 120 numbers into groups based on their first digit, you'll find there are 120/5 = 24 numbers in each group starting with a 1,3,5,7 or 9.

So since there are 24 numbers in the 1nnnn group and 24 numbers in the 3nnnn group, we know that the 60th number must be in the 5nnnn group.

Now this is where Jim left off, but it turns out we can also use Jim's logic to continue on to find the second digit in the 5nnnn group:

Since we know that the 5-digit number we want starts with (and uses up) the digit "5", then we also know there are 4*3*2 = 24 ways of arranging the remaining digits: 1,3,7,9

And just as before, if you divide these 24 numbers into groups based on their first digit, you'll find there are 24/4 = 6 numbers in each group starting with a 1,3,7 or 9.

So since there are 6 numbers in the 51nnn group and 6 numbers in the 53nnn group, and that 24+24+6+6=60, then we know that the 60th number must be the last number in the 53nnn group, which would make it 53971.

But since the median is the "middle" value in a sequence, the "middle" of a list of 120 numbers lies between the 60th and the 61st number. This means the median must be artificially created by averaging the 60th and 61st number.

So since the 60th number was the LAST number in the 53nnn group, then the 61st number must be the FIRST number in the following group which is 57nnn, and the first number in that group would be 57139.

1 13579
2 13597
3 13759
.. ...
56 ...
57 53719
58 53791
59 53917
60 53971 - last number in the 53nnn group
60.5 - median position in a list of 120 numbers
61 57139 - first number in the 57nnn group
62 57193
63 57319
64 57391
65 ...
.. ...
118 97351
119 97513
120 97531

So bottom line here is that I figure the median would be (53971+57139)/2

Numbers check anyone? :-)

Answer 17

that's clever, didn't think to extend it but that makes sense now that I think about it

and yes, the median is 55,555

Answer 18

YAY!!, as they say... :-)

And thank you!

Answer 19

you're welcome