Kamil wrote on a board an expression consisting of alternating addition and subtraction of natural numbers from $1$ to $100$: \[1-2+3-4+5-6+\ldots-98+99-100.\]Then, Kamil erased one of the plus or minus signs and replaced it with an equals sign, obtaining a true equality. Which number preceded the erased sign? Find all possibilities and justify your answer.
Problem
Source:
Tags: combinatorics
Awesome_Twin1
21.05.2023 22:26
All of the posts you made about this contest should be moved to the High School Olympiads if it is high school level or Contests and Programs. \req move to C and P.
FarhanFeroz
29.08.2023 13:09
Kamil has two options: Option 1: Replacing a plus sign with an equals sign. In this case, the number of terms on each side of the equal sign is even. And every consecutive pair of the series gives a sum of -1. So if he replaces any plus sign having n terms on the left and m terms on the right. Then, sum on the left side = -(n/2) and sum on the right side = -(m/2). Since, he obtains a true equality, -(n/2) = -(m/2) i.e. n=m and n+m=100 Therefore, n = m = 50. So in this case this is the only possibility and we will have the 50th term that is 50 preceding the erased sign. Option 2: Replacing a negative sign with an equal sign. In this case, we will have an odd number of terms on both sides of the equal sign. Let x be the number preceding the erased sign, thus the number following the erased sign will be (x+1). Except for these two numbers, all other remaining numbers will have their pair on their respective side to give a sum of -1. If there are n terms on the left side before x and m terms on the right side after (x+1). Then we will get, -(n/2) + x = (x+1) - (m/2) i.e. -(n/2) = 1 - (m/2) and this time excluding x and (x+1) we have, n+m = 98. Solving these two equations we have a unique solution, n = 48 and m = 50. That means x is the 49th term that is we will have 49 preceding the erased sign. These are the only two possibilities.