Grade $6$ Math Problem 
$10$ players are playing in a card game in which the winner is the player having the most number of cards. There are $230$ cards in total. What is the smallest number of cards the winning player could have collected, assuming that each player collected a different number of cards? 

When I attempted the question, I tried the highest consecutive numbers the other $9$ players could have and then added them and took that number away from $230$. 
So 
$$26+25+24+ \cdots + 18 = 198$$
The winner has then $230-198= 32$ cards.
 A: If the first player had $27$ or less cards, then the maximum total number of cards possible is $27+26+25+\cdots+18=\frac{27\times28}{2}-\frac{17\times18}{2}=225$ cards, so the first player must've had at least $28$ cards.
On the other hand, note that $28+27+26+25+24+22+21+20+19+18=230$ so the first player having $28$ cards is completely possible - and thus $28$ is the answer.
edit: Just a note (reading comments on the main post) - this assumes that every card was collected by a player - but this is kinda implicit in the question, otherwise the first player could've just collected $9$ cards with everybody else collecting $8,7,6,5,4,3,2,1,0$ which is kinda trivial.
A: I'll attempt to find a closed form solution to the problem. Let $n$ be the number of players and $c$ be the total number of cards. The first player must have at least $u+1$ cards, where $u$ is the highest integer that satisfies
$$\sum_{k=1}^{n}k+u<c$$
However, this sum can also be expressed as $$un+\sum_{k=1}^{n}k=un+\frac{1}{2}n(n+1)=n\left (u+\frac{n+1}{2}\right )$$
And therefore
$$n\left (u+\frac{n+1}{2}\right )<c\implies u<\frac{c}{n}-\frac{n+1}{2}$$
$$u<\frac{2c-n(n+1)}{2n}$$
So the least integer that satisfies these requirements is $$N=\lceil\frac{2c-n(n+1)}{2n}+1\rceil=\lceil\frac{2c-n(n-1)}{2n}\rceil$$
$$N=\lceil \frac c n+\frac{n-1}{2}\rceil$$
In your particular case, $$N=\lceil \frac{230}{10} + \frac{9}{2}\rceil=\lceil 27.5\rceil=28$$
A: Extending this to $n$ cards and $k$ players, let $m$ be the minimum possible sum with $k$ players i.e. $m=\sum_{i=1}^ki$. The solution to the problem is then $\lceil \frac {n-m}k \rceil+k$.
e.g. $n=230$, $k=10$, then $m=55$. Solutions is:
$\lceil \frac {230-55}{10} \rceil+10=\lceil \frac {175}{10} \rceil+10=\lceil 17.5 \rceil+10=17+10=28$
Proof: $28+27+26+25+24+22+21+20+19+18=230$
e.g. $n=21$, $k=4$, then $m=10$. Solutions is:
$\lceil \frac {21-10}{4} \rceil+4=\lceil \frac {11}{4} \rceil+4=\lceil 2.75 \rceil+4=3+4=7$
Proof: $3+5+6+7=21$
e.g. $n=23$, $k=4$, then $m=10$. Solutions is:
$\lceil \frac {23-10}{4} \rceil+4=\lceil \frac {13}{4} \rceil+4=\lceil 3.25 \rceil+4=4+4=8$
Proof: $4+5+6+8=23$
