Flipping Signs in 1+3+5+…+99

🧭 Problem

In the expression \(1+3+5+7+\cdots+97+99\), A boy changes some plus signs to minus signs. The new value is negative. What is the least number of plus signs she could have changed?

(A) 14 (B) 15 (C) 16 (D) 17 (E) 18

1️⃣ Evaluate the original total

There are \(50\) odd terms from \(1\) to \(99\). Using the classic identity:

\[ 1+3+\cdots+(2n-1)=n^2 \quad\Rightarrow\quad 1+3+\cdots+99 = 50^2 = 2500. \]

2️⃣ Make it most negative with the fewest flips

To minimize the number of flips, change the largest terms to negative. Keep the first \(n\) odds positive and flip the rest:

\[ \underbrace{(1+3+\cdots+(2n-1))}_{\text{kept positive}} \;-\; \underbrace{((2n+1)+(2n+3)+\cdots+99)}_{\text{flipped}} \;<\;0. \]

The positive part sums to \(n^2\). The flipped part is the remainder, with sum \(2500-n^2\). So the new value is

\[ n^2-(2500-n^2)=2n^2-2500. \]

We want it negative:

\[ 2n^2-2500<0 \;\Longrightarrow\; n^2<1250 \;\Longrightarrow\; n\le 35. \]

3️⃣ Count the flips

With \(n=35\) positives, the number of flipped signs is the remaining odd terms:

\[ 50-35=15. \]

If \(n=36\), then \(2n^2-2500=2592-2500=92>0\) (not negative). Hence the least number of flips is \(15\) → option (B).

📝 Quick Check

Same idea for \(1+3+\cdots+199\) (there are \(100\) terms). What is the least number of flips to make it negative?

← Prev: Cyclic Quadrilateral — Shorter Diagonal Index: CAT Quant Cheat Codes Next: Log Puzzle: (log₂x · log₃x)/(log₂x + log₃x)

Flipping Signs in 1+3+5+…+99

Flip Some Signs in \(1+3+5+\cdots+99\) to Make the Value Negative

🧭 Problem

1️⃣ Evaluate the original total

2️⃣ Make it most negative with the fewest flips

3️⃣ Count the flips

📝 Quick Check

Address

Explore