What is the total number of positive integers <300 whose digit sum is a multiple of 5? As stated in the title
For one digit number there is $1$
For two digit numbers there are $27(=4+(4+5)+(4+5+5)) (sum
=5n, n=1,2,3)$ $since (5n<18)$
But I'm struggling to find the ones that are of three digits 
$(sum)=5n, n=1,...,4)$$for (5n\leq20)$ (without writing out all the possible combinations).
The possible answers are $18,45,60,61$
Can someone please explain which of the above answers is the most appropriate?
 A: If the first two digits are $0,0$ than the last digit must be $0$ or $5$. There are two such numbers.
If the first two digits are $0,1$ then the last digit must be $4$ or $9$.  There are two such number.
If the first two digits are $2$ and $7$ that then last digit must be $1$ and $6$.  There are two such numbers.
And so on.
The first two digits are $00$ to $29$, that's $30$ option and there are two options for $c$ so that there are $2*30 = 60$ such numbers.  But $000$ is not acceptable so there are $59$.
So, $59$.  Final answer.
....
But we must prove that if the first two digits are $a$ and $b$ there are exactly $2$ digits that third digit can be.
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ANd... if you add up $a$ and $b$ and take the remainder from dividing be $5$ (call it $r$) there are exactly $5$ possible remainders values of $r$.  For $a+b +c$ to be a multiple of $5$ then the remainder $a+b+c$ must be $0$ and the remainder of $c$ must be the opposite $r$.  That is if $r=1$ then the remainder of $c$ (call it $s$) must be $4$ and if $r=2$ then $s = 3$ and $r=3$ then $s=2$ and if $r=4$ then $s=1$ and if $r=0$ then $s = 0$.
And there are exactly $5$ possible opposite remainders that $s$ can be.  $c$ can be $s+0$ and $s+5$.
....
Putting that argument in terms of
Modular arithmetic:  If $a+b \equiv r \pmod 5$ then $c \equiv - r\pmod 5$.  As $\{0,1,2,3,4\}$ and $\{5,6,7,8,9\}$ are two complete residue systems there are exactly one representative for $-r$ in each set so there are two option for digit $c$.
.....
Remainder theorem:
Let $a+b = 5K + r$.  $a+b+c = 5M + 0$.  Then $c= (5M + 0)- (5K+r)=5(M-K)+r$ and $0\le c \le 9$ and $0 \le r \le 4$.  If $r=0$ then $0\le c= 5(M-K)\le 9$ so $c=0, 5$.  If $0 < r < 5$ then $0\le c = 5(M-k)-r \le 9$ then $c = 5-r$ or $c = 10-r$.
A: Two hints:


*

*Pad any 1 or 2-digit numbers with leading zeroes so that you need only consider 3-digit numbers.

*If you are given the first two digits of a 3-digit number, and you know the number has a digit sum that is a multiple of 5, then how many choices do you have left for the last digit?
Note that the numbers in the question should probably range from 000 to 299.
A: I wasn't able to follow the OP's reasoning re there being 27 such two digit numbers.  Also, I was surprised by the offered answers, because I think that the answer is 59.  I infer this because if the numbers are allowed to range from 0 thru 299, rather than 1 thru 299, then I would say that the answer is 60 = 300/5.
Initially, when I considered this problem, I considered that the sum of the digits in 19 = 10 (a multiple of 5), but the sum of the digits in 20 [= 19 + 1] is not congruent mod 5 to [10 + 1].
Then, I realized that regardless of what the sum of the digits of 20 is [mod 5], in the range 20 thru 29, there must be exactly two numbers the sum of whose digits is a multiple of 5.  The same can be said for the 10 numbers 30 thru 39, 40 thru 49, ..., 100 thru 109, ...
This is why I conclude that if the numbers are allowed to range from 0 thru 299, there must be 60 such numbers.
