AMC 10 · 2024 · #5

Grade 6 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target 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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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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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