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.

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: $4+5+6+8=23$
