AMC 10 · 2024 · #5

Grade 6 algebra

Archetype Small-Domain Constrained Search

sequences-arithmeticperfect-squaressystematic-enumeration convert-to-algebrabound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: sequences-arithmeticlinear-equations-one-var

📏 Medium solution 💡 3 insights

Problem

In the following expression, Melanie changed some of the plus signs to minus signs:
$1+3+5+7+...+97+99$
When the new expression was evaluated, it was negative. What is the least number of plus signs that Melanie could have changed to minus signs?

$\textbf{(A) } 14 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 16 \qquad\textbf{(D) } 17 \qquad\textbf{(E) } 18$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Start with the expression $1 + 3 + 5 + \cdots + 97 + 99$ (the sum of the first $50$ positive odd integers). Melanie flips some of the $+$ signs to $-$ signs so that the new value is negative. What is the smallest number of sign flips that can achieve this?

Givens: Original expression: $1 + 3 + 5 + \cdots + 97 + 99$, with $50$ terms; Original sum $S = 1 + 3 + \cdots + 99 = 50^2 = 2500$; Each sign flip on the number $x$ changes the total by $-2x$ (from $+x$ to $-x$); Answer choices: (A) $14$, (B) $15$, (C) $16$, (D) $17$, (E) $18$

Unknowns: The minimum number of $+$ signs Melanie must change to make the new sum negative

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #6 Guess and Check, #16 Change Focus / Count the Complement

Tool #13 (Convert to Algebra) turns the verbal "flip plus to minus" rule into clean arithmetic: each flip of $x$ subtracts $2x$ from the original total $S = 2500$, so the new sum is $S - 2 \cdot (\text{sum of flipped terms})$ and we need this $< 0$, i.e. the flipped terms must sum to more than $1250$. Tool #16 (Change Focus) is the strategic insight that lets us minimize the count — instead of asking "how many flips?", ask "what's the maximum reduction per flip?" and you get the greedy choice $99, 97, 95, \ldots$. Tool #6 (Guess and Check) finishes by testing the two adjacent candidate values $k = 14$ and $k = 15$ in the closed-form sum.

Execute — Answer: B

#13 Convert to Algebra 6.EE.A.3 Step 1

Compute the original sum.
The first $50$ positive odd integers add to $50^2 = 2500$ — a standard identity ($1 + 3 + \cdots + (2n-1) = n^2$) that can be checked by pairing terms: $1 + 99 = 100$, $3 + 97 = 100$, $\ldots$, $25$ pairs of $100 = 2500$.

$$S = 1 + 3 + 5 + \cdots + 99 = 50^2 = 2500$$

💡 Pairing the first and last terms ($1 + 99 = 100$) and walking inward is the same equivalent-expressions trick Grade 6 introduces for arithmetic sums.

#13 Convert to Algebra 6.EE.A.2 Step 2

Quantify the effect of a single sign flip.
Changing the term $+x$ to $-x$ moves the sum down by $(-x) - (+x) = -2x$.
If Melanie flips a set of numbers whose total is $F$, the new sum is $2500 - 2F$.

$$S_{\text{new}} = 2500 - 2F, \quad F = \sum_{\text{flipped}} x$$

💡 Letting $F$ stand for "sum of flipped numbers" turns the verbal rule into a clean expression — the Grade 6 "letters stand for numbers" idea.

#13 Convert to Algebra 6.EE.B.8 Step 3

Write the "new sum is negative" condition as an inequality on $F$.

$$2500 - 2F < 0 \;\Longleftrightarrow\; F > 1250$$

💡 Solving "strictly negative" for the flipped-sum gives the inequality $F > 1250$ — a Grade 6 "write/solve inequality of the form $x > c$" step.

#16 Change Focus / Count the Complement 6.EE.A.3 Step 4

Change focus: to minimize the number of flips, each flip should pull the most weight.
So greedily flip the largest odd numbers $99, 97, 95, \ldots$.
The $k$ largest odd numbers form an arithmetic sequence from $101 - 2k$ up to $99$, summing to $\tfrac{k}{2}(99 + (101 - 2k)) = k(100 - k)$.

$$\text{max sum with } k \text{ flips} = 99 + 97 + \cdots + (101 - 2k) = k(100 - k)$$

💡 Picking the heaviest numbers first is the "do the most with the fewest" extreme-principle move; the arithmetic-series formula is a Grade 6 equivalent-expressions calculation.

#6 Guess and Check 5.NBT.B.5 Step 5

Find the smallest $k$ with $k(100 - k) > 1250$ by checking the two candidate answers near the boundary.

$k = 14: \; 14 \cdot 86 = 1204 \not> 1250 \quad\text{(fails)}$ $\qquad$ $k = 15: \; 15 \cdot 85 = 1275 > 1250 \quad\text{(works)} \;\Rightarrow\; \textbf{(B)}$

💡 Two two-digit multiplications bracket the threshold cleanly — Grade 5 multi-digit multiplication is enough.

[1] #13 6.EE.A.3 Compute the original sum. The first $50$ positive odd integers add to $50^2 = 25

[2] #13 6.EE.A.2 Quantify the effect of a single sign flip. Changing the term $+x$ to $-x$ moves

[3] #13 6.EE.B.8 Write the "new sum is negative" condition as an inequality on $F$.

[4] #16 6.EE.A.3 Change focus: to minimize the number of flips, each flip should pull the most we

[5] #6 5.NBT.B.5 Find the smallest $k$ with $k(100 - k) > 1250$ by checking the two candidate ans

Review

Reasonableness: Verify the $k = 15$ flip explicitly. Flipping the top $15$ odd numbers $71, 73, \ldots, 99$ pulls them off the positive side and adds them on the negative side, costing $2 \cdot 1275 = 2550$. The new sum is $2500 - 2550 = -50 < 0$, so $15$ flips really do produce a negative total. With $k = 14$, even the most efficient flips ($73, 75, \ldots, 99$) sum to $1204$, giving $2500 - 2408 = 92 > 0$, so $14$ flips cannot do it. The minimum is exactly $15$, matching (B).

Alternative: Tool #6 (Guess and Check) without the closed-form sum: directly add the largest odd numbers $99 + 97 + 95 + \cdots$ until the running total crosses $1250$. Running cumulative sums: after $14$ numbers (down to $73$) the total is $1204 < 1250$; adding $71$ gives $1275 > 1250$. So at the $15$th flip the crossing happens — answer (B).

CCSS standards used (min grade 6)

5.NBT.B.5 Fluently multiply multi-digit whole numbers (Computing $14 \cdot 86 = 1204$ and $15 \cdot 85 = 1275$ to test which is the smallest $k$ with $k(100 - k) > 1250$.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Letting $F$ stand for the sum of flipped numbers and writing the new total as $2500 - 2F$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Computing $S = 50^2 = 2500$ via pairing $1 + 99, 3 + 97, \ldots$ and deriving $99 + 97 + \cdots + (101 - 2k) = k(100 - k)$.)
6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Turning the condition "new sum is negative" into the inequality $F > 1250$ for the sum of flipped numbers.)

⭐ This AMC 10 problem only needs Grade 6 inequalities (turn "new sum negative" into $F > 1250$) plus Grade 5 multi-digit multiplication — flip the biggest odd numbers first and you cross the line at flip $15$!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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