Question

How many three-digit numbers are there in which the sum of the digits is even?

Answer 1

This is an interesting problem which can be tackled using any of
several different approaches. One very good approach is to look at
the numbers of combinations of even digits, as you seem to suggest.

We want the number of even digits to be even, so we can have either 0
or 2 even digits in our 3-digit number. If we represent even digits
with E and odd digits with O, then we can have 3-digit numbers of any
of these forms:

EEO
EOE
OEE
OOO

We can find the numbers of different 3-digit numbers of each of these
forms using the fundamental counting principle. In doing this, we
need to remember that we have 5 choices for each even or odd digit,
except in the case of a leading even digit, where we have only 4
choices, since a leading digit of 0 is not allowed. So we have

EEO: 4*5*5 = 100 3-digit numbers of this form
EOE: 4*5*5 = 100 3-digit numbers of this form
OEE: 5*5*5 = 125 3-digit numbers of this form
OOO: 5*5*5 = 125 3-digit numbers of this form
-----------------------------------------------
total: 450

So there are 450 3-digit numbers with an even number of even digits.

To give us some confidence in our method, we can use the same process
to find the number of 3-digit numbers with an odd number of even
digits. Together with the number of 3-digit numbers with an even
number of even digits, this should give us a grand total of 900, which
is the total number of 3-digit numbers.

EEE: 4*5*5 = 100 3-digit numbers of this form
EOO: 4*5*5 = 100 3-digit numbers of this form
OEO: 5*5*5 = 125 3-digit numbers of this form
OOE: 5*5*5 = 125 3-digit numbers of this form

This total is also 450, giving us the required grand total of 900 3-
digit numbers, so our method and our computations are probably
correct.

Note that the numbers of 3-digit numbers with an even number of even
digits and with an odd number of even digits are the same--450 each.
There is a very sophisticated method for solving this problem very
quickly using this fact; we show that exactly half of all 3-digit
numbers have an even number of even digits, so the number of 3-digit
numbers with an even number of even digits is half of 900, or 450.

Following is an informal argument for why exactly half of all 3-digit
numbers have an even number of even digits:

Consider a particular 3-digit number, say 384. Find the number which
is the "9's complement" of that number--that is, the number which when
added to the given number gives a result of 999. For this example, we
have 999-384=615. So 384 and 615 are a "pair" of numbers related by
the special fact that their sum is 999.

The original number contains two even digits. We found the other
number by subtracting the original number from 999. 9 is an odd
digit; and an odd digit can be expressed as the sum of two digits only
as the sum of one odd and one even digit. Therefore, every even digit
in our original number will correspond to an odd digit in our "new"
number, while every odd digit in our original number will correspond
to an even digit in our "new" number:

3 <==> 6
8 <==> 1
4 <==> 5

We can think of pairing up all 900 3-digit numbers with their "9's
complement" numbers in this manner. The numbers all contain 3 digits;
and in each pair, each even digit of one number corresponds to an odd
digit of the other number and vice versa. Therefore, in each pair,
one of the two 3-digit numbers will have an even number of even digits
and the other will have an odd number of even digits. And since our
complete list of pairings contains all 900 3-digit numbers, exactly
half of those 900 3-digit numbers have an even number of even digits
and exactly half have an odd number of even digits.

I hope all this helps. Please write back if you have any further
questions about any of this.

Answer 2

Lol, guy, i looked at this. Its a slightly different problem.

Answer 3

A small Mathematica program solution is attached.

Answer 4

Rob, thank you! But I have found out the solution already, and it isn't 50. This one is a real thinker. If you think about it, you can only have a certain amount of Even and Odd combinations. Which are: OOE, OEO, EOO, and EEE.
Using these combinations, we can say that we have 2(5*5*5)+2(4*5*5)=450 different combinations. The first part comes from (OOE, and OEO). The second part comes from (EEE and EOE), but we multiply by 4 and not 5 because the first number cannot be zero.

Rob, I do have a question for you though, are you proficient with Mathematica? Because I am doing research using mathematica and would like some help using it.

Answer 5

I just realized you used 100-199, so essentially, that is right when you go from 100-999. That is one impressive line of code :)

Answer 6

I did not log out last night and have some things to do this morning, California time. Will respond later today.