Problem 4 · A hundred and one terms centred at k
The symmetric sum equals $101k$, and $10201 = 101^2$.
Integer answer, at most 4 digitsGiven that $(k-50) + (k-49) + \ldots + (k-1) + k + (k+1) + \ldots + (k+50) = 10201$, what is $k$?
Copa Cangur · SCM
Easy
Closed answer
Reasoned solution
Key idea: the shifts $-50, \ldots, +50$ cancel in symmetric pairs.
There are $101$ terms and the shifts sum to zero, so all that remains is:
$$101k = 10201 = 101^2 \;\Longrightarrow\; k = 101.$$
Answer: 101