# 3-digit numbers that the sum of digits are even.

How many three digit numbers are there such that the sum of the digits is even?

So I guess we're taking the total number of three digit numbers, then eliminate the ones that doesn't satisfy the properties. But, can someone give me a hint on how to count the number of 3-digit numbers that has a sum of digits that's even?

Ty!

• If I give you any two digits X and Y, what digit Z can you give me back so that X+Y+Z is even? Commented Aug 3, 2013 at 16:55
• Why would it be easier to count the ones with odd sums? Commented Aug 3, 2013 at 16:59
• Basically, given the first two digits, exactly have of $\{0,\dots,9\}$ can be used for the last digit. Commented Aug 3, 2013 at 16:59

HINT: Suppose that the digits are $abc$.

• How many choices are there for $ab$?
• How many choices are there for $c$ if $a+b$ is even? How many choices are there for $c$ if $a+b$ is odd? Does it make any difference whether $a+b$ is odd or even?

A simple way of seeing that it is exactly half, is to realize that we have a bijection between 3 digit numbers by sending $N$ to $1099- N$. Note that since the sum of these numbers is odd, hence the pair of numbers have different parity.

Put explicitly, we are pairing up $\{100, 999\}, \{101, 998\} \ldots, \{544, 545 \}$.

Ways to get an even sum:

$2$ odd $1$ even.

$3$ even.

First case:

First see that there are three posiblities for the place the even number takes: (the first number cant be 0 because then it would not be a 3 digit number. Therefore there are $(4*5*5)+2(5*5*5)$ ways

second case: all even. there are $4*5*5$ ways to do it therefore the total is $100+250+100=450$ numbers

Exactly half of them.

If you allow leading zeroes, there are $1000$ three-digit numbers, if you are strict, there are $900$ three-digit numbers, so the final answer is $500$ or $450$, depending on your definition. Note that among any two consecutive three-digit numbers tha tdiffer only in their last digit, one has even and one has odd digit sum.

• So, if we're counting for 4-digit numbers, or 5-digit, so-on, it's still half of them that have a digit sum that's even/odd? Commented Aug 3, 2013 at 16:57
• This is not a hint. Commented Aug 3, 2013 at 16:57
• I'm not sure for your result: notice that we begin by 100 and end by 999 so I think that the number is the half-1=449. Isn't it?
– user63181
Commented Aug 3, 2013 at 17:10
• no. substract 99 to all terms to get {1,2,3...900} which clearly has 900 elements Commented Aug 3, 2013 at 17:13

HINT If the final digit of $n$ (units digit) is not $9$ then $n$ and $n+1$ have one odd digit sum and one even sum. Can you see how to pair numbers up so that one is odd and one is even, and every three-digit number is included in the pairing?

HINT:

So, the number digits with odd value is even

If we take no digits with odd values, we shall have $4\cdot 5\cdot5=100$ combinations ,

If we take two digits with odd values,

we have following combinations $(O,O,E),(O,E,O),(E,O,O)$

Now, for the first digit in even case, it can assume $4$ values namely, $\{2,4,6,8\}$

and in odd case, it can assume, $5$ values namely, $\{1,3,5,7,9\}$

• {0,2,4,6,8} has 5 elements Commented Aug 3, 2013 at 17:02
• @Omnitic, the first digit can not be $0,$ for a valid three digit number, right? Commented Aug 3, 2013 at 17:03
• should it not be 4*5*5 then? Commented Aug 3, 2013 at 17:04
• @Omnitic, sorry, Rectified. I wrongly included one extra condition: without repetition Commented Aug 3, 2013 at 17:06

Since the even number = sum 3 even or sum 2 even, 1 odd so we have 4 case:

1. The number is $$\, \overline{abc}\,$$ where $$a,b,c\in \{ 0,2,4,6,8\}$$. The number $$a$$ has $$4$$ choices, the number $$b$$ has $$4$$ choices, the number $$c$$ has $$3$$ choices. Therefore, we have $$\, 4*4*3=48,$$ numbers.
2. The number is $$\, \overline{abc}\,$$ where $$a\in \{ 2,4,6,8\}$$ and $$b,c \in \{1,3,5,7,9 \}$$. The number $$a$$ has $$4$$ choices, the number $$b$$ has $$5$$ choices, the number $$c$$ has $$4$$ choices. Therefore, we have $$\, 4*5*4=80\,$$ numbers.
3. The number is $$\, \overline{abc}\,$$ where $$a,c\in \{ 0,2,4,6,8\}$$ and $$b\in \{1,3,5,7,9\}$$. The number $$a$$ has $$4$$ choices, the number $$c$$ has $$4$$ choices, the number $$b$$ has $$5$$ choices. Therefore, we have $$\, 4*4*5=80\,$$ numbers.
4. The number is $$\, \overline{abc}\,$$ where $$c\in \{ 0,2,4,6,8\}$$ and $$a, b\in \{1,3,5,7,9\}$$. The number $$a$$ has $$5$$ choices, the number $$b$$ has $$4$$ choices, the number $$c$$ has $$5$$ choices. Therefore, we have $$\, 5*4*5=100\,$$ numbers. The conclution, we have $$\, 48+80+80+100=308$$ numbers.