Question

A combinatorics problem!

Prove that the number of the 4-digits numbers that the sum of their digits is a even number is equal to the number of the 4-digits numbers which the sum of their digits is an odd number.

Answer 1

Oh hold on, I think I follow...

Answer 2

sry for grammer...yes translating these sorts of problems is really difficult :) imagine we have 2222 , it's a 4 digits number and the sum of their digits (2+2+2+2) = 8 and it's a even number...and for 3233 we have 3+2+3+3 = 11 and it's an odd number

Answer 3

The use of "number" and "numbers" was a bit confusing at first

Answer 4

and the question wants us to prove that the numbers of these numbers are equal to each other...

Answer 5

if u have understood the problem then just please write the correct question for me to edit it... ;)

Answer 6

Yeah I'm trying to express the question in a slightly different and perhaps rigorous way...

Answer 7

but i think the grammer is correct ;)

Answer 8

maybe the explanation isn't very good.

Answer 9

The set of all 4-digit numbers whose digits add up to an even number could be written as
\[A=\left\{1000x_4+100x_3+10x_2+x_1~\bigg|~x_1,x_2,x_3,x_4\in\{0,1,...,9\},~\sum_{i=1}^4x_i=2k,~k\in\mathbb{N}\right\}\]
Is that correct?

Answer 10

yes

Answer 11

The cut-off part is \(k\in\mathbb{N}\), i.e. \(k\) in a natural number

Answer 12

Okay so \(B\) would have the same basic definition, with \(2k+1\) instead of \(2k\).

Answer 13

yes i know what does 2k mean ;)

Answer 14

but i don't think it's a strong logic

Answer 15

Case by case:
\[(1)~\sum_{i=1}^4x_i~\text{is even}\]
Then (a) all \(x_i\) are even, (b) only two \(x_i\) are even, or (c) all \(x_i\) are odd.

You should be able to easily count how many 4-digit numbers you can have under each sub-case.

Answer 16

you mean i can use :

and simply say |A| = |B| ?

Answer 17

No that's not what I was getting at. I have a different systematic method.

For example under (a), you can have for any given number in \(A\)
\[\underbrace{(2,4,6,8,0)}_{5\text{ choices}}\times\underbrace{(2,4,6,8,0)}_{5\text{ choices}}\times\underbrace{(2,4,6,8,0)}_{5\text{ choices}}\times\underbrace{(2,4,6,8,0)}_{5\text{ choices}}=5^4\text{ choices}\]

Answer 18

You'd get the same number under (b) and (c), so it looks like for case (1) you have \(3(5^4)\) numbers.

Answer 19

oh wait!

Answer 20

i know what you are going to say! u want to find the number of these numbers and simply subtract from (9x10x10x10) ?

Answer 21

Essentially, yes.

I have to point out a mistake though. The first number can't have a 0, so there should be
\[\underbrace{(2,4,6,8)}_{5\text{ choices}}\times\underbrace{(2,4,6,8,0)}_{5\text{ choices}}\times\underbrace{(2,4,6,8,0)}_{5\text{ choices}}\times\underbrace{(2,4,6,8,0)}_{5\text{ choices}}=4(5^3)\text{ choices}\]
So \(A\) should have \(12(5^3)\) numbers.

Answer 22

I think the question boils down to:

For numbers 1000 to 9999, when individual digits are added
how many of them will be even and how many will be odd?

1+0+0+0 is odd
1+0+0+1 is even

9+9+9+8 is odd
9+9+9+9 is even

As many even as there are odd.

Answer 23

Yeah that's a much simpler approach!

Answer 24

@SithsAndGiggles , i know how do the rest ;) thank you...really this post helped me alot:

\( \large \color{lime}{A=\left\{1000x_4+100x_3+10x_2+x_1~\bigg|~x_1,x_2,x_3,x_4\in\{0,1,...,9\},~\sum_{i=1}^4x_i=2k,~k\in\mathbb{N}\right\}} \)

Answer 25

thank you...i just translated it from my book :)

Answer 26

No problem

Answer 27

Just realized my solution won't work!

1009: sum of digits is even
The next number 1010: sum of digits is also even!

So the sums of digits of consecutive integers don't necessarily alternate.

Answer 28

It looks like the pattern would be
\[\underbrace{a~b~a~b~\cdots~a~b}_{1000-1009}~\underbrace{b~a~b~a~\cdots~b~a}_{1010-1019}~a~b~a~b~\cdots\]
where \(a\) and \(b\)'s digits add up to an odd and end even number, respectively. So, the method could still work.