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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!

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  • $\begingroup$ 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? $\endgroup$
    – RghtHndSd
    Commented Aug 3, 2013 at 16:55
  • $\begingroup$ Why would it be easier to count the ones with odd sums? $\endgroup$
    – Thomas Andrews
    Commented Aug 3, 2013 at 16:59
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    $\begingroup$ Basically, given the first two digits, exactly have of $\{0,\dots,9\}$ can be used for the last digit. $\endgroup$
    – Thomas Andrews
    Commented Aug 3, 2013 at 16:59

7 Answers 7

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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?
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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 \} $.

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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

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Commander Shepard
    Commented Aug 3, 2013 at 16:57
  • $\begingroup$ This is not a hint. $\endgroup$
    – RghtHndSd
    Commented Aug 3, 2013 at 16:57
  • $\begingroup$ 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? $\endgroup$
    – user63181
    Commented Aug 3, 2013 at 17:10
  • $\begingroup$ no. substract 99 to all terms to get {1,2,3...900} which clearly has 900 elements $\endgroup$
    – Asinomás
    Commented Aug 3, 2013 at 17:13
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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?

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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, $\{1,3,5,7,9\}$

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