AMC 10 · 2019 · #5

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition

sequences-arithmeticsymmetry-argumentpattern-recognitionmean-median-mode-range pattern-recognitionidentify-subproblems ↑ Prerequisites: sequences-arithmetic

📏 Short solution 💡 2 insights

Problem

What is the greatest number of consecutive integers whose sum is $45?$

$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

120

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Toolkit + CCSS Solution

Understand

Restated: Find the largest number of consecutive integers (positive, negative, or zero allowed) whose sum is exactly $45$.

Givens: The sum of the chosen consecutive integers must equal $45$; The integers must be consecutive — no gaps allowed; Negative integers and zero are permitted (the problem says 'integers', not 'positive integers'); Answer choices: (A) $9$, (B) $25$, (C) $45$, (D) $90$, (E) $120$

Unknowns: The maximum count of consecutive integers in such a sequence

Understand

Restated: Find the largest number of consecutive integers (positive, negative, or zero allowed) whose sum is exactly $45$.

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #1 Draw a Diagram, #5 Look for a Pattern, #3 Eliminate Possibilities

Tool #9 (Easier Problem): try a smaller target sum first, like 'consecutive integers summing to $5$'. The trick of including negatives so most terms cancel pops out: $-4, -3, \ldots, 4, 5$ sums to $5$ and has $10$ terms. Tool #1 (Diagram): a number line shows the symmetric block from $-k$ to $+k$ summing to $0$, with a tail $\{k+1, \ldots, 45\}$ paying for the $45$. Tool #5 (Pattern): once we see the smaller case, generalize — use $-44$ to $45$ for sum $45$, giving $90$ terms. Tool #3 (Eliminate): the choices climb $9, 25, 45, 90, 120$. Anything $\le 45$ ignores negatives. $120$ is too long because the symmetric trick caps at $90$.

Execute — Answer: D

#9 Solve an Easier Related Problem 6.NS.C.6 Step 1

Warm up with a smaller target.
How many consecutive integers can sum to $5$?
Use $-4, -3, -2, -1, 0, 1, 2, 3, 4, 5$.
The pairs $(-4, 4), (-3, 3), (-2, 2), (-1, 1)$ cancel; $0$ contributes nothing; only $5$ remains.
The count is $10$ terms — much more than the obvious $\{5\}$ alone or $\{2, 3\}$.

$(-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 = 5$, with $10$ terms

💡 Negatives cancel positives in pairs — the only 'survivor' is the top number.

#5 Look for a Pattern 6.NS.C.6 Step 2

Generalize the trick.
For a target sum $S > 0$, take the integers from $-(S - 1)$ up to $S$.
Pairs $(-k, k)$ for $k = 1, 2, \ldots, S - 1$ cancel, and the middle $0$ adds nothing, so the sum is just $S$.
The count goes from $-(S - 1)$ to $S$, which is $S - (-(S-1)) + 1 = 2S$ terms.

Integers $-(S-1), -(S-2), \ldots, -1, 0, 1, \ldots, S - 1, S$ — count $= 2S$, sum $= S$

💡 Pattern from the warm-up — sum survives, length doubles.

#5 Look for a Pattern 6.NS.C.6 Step 3

Apply to $S = 45$.
Take the integers from $-44$ up to $45$.
That is $-44, -43, \ldots, -1, 0, 1, \ldots, 44, 45$.
All pairs $(-k, k)$ cancel for $k = 1, \ldots, 44$, the $0$ adds nothing, and only $45$ remains.
The count is $45 - (-44) + 1 = 90$.

Sequence $-44, -43, \ldots, 44, 45$;\ count $= 45 - (-44) + 1 = 90$;\ sum $= 45 \;\Rightarrow\; \textbf{(D)}$

💡 Same pattern as the warm-up, just bigger: from $-(S-1)$ to $S$ gives $2S$ terms.

#9 Solve an Easier Related Problem 6.EE.B.7 Step 4

Confirm this is the maximum.
A block of $n$ consecutive integers starting at $a$ sums to $n \cdot a + \frac{n(n-1)}{2}$.
Setting this equal to $45$ and clearing fractions: $2na + n(n-1) = 90$, i.e.
$n \big( 2a + (n - 1) \big) = 90$.
So $n$ must divide $90$.
The largest divisor of $90$ in the choice list is $90$ itself (and we already exhibited it), so $n = 90$ is optimal.

$n \big( 2a + n - 1 \big) = 90 \;\Rightarrow\; n \mid 90;$ largest divisor $\le 120$ is $90$, achieved $\;\Rightarrow\; \textbf{(D)}$

💡 Length must divide $90$ — and $90$ itself works.

#3 Eliminate Possibilities 4.OA.B.4 Step 5

Eliminate the smaller choices.
(A) $9$, (B) $25$, (C) $45$ are all divisors of $90$ that work too, but they aren't the maximum — we already beat them.
(E) $120$ fails because $120 \nmid 90$ (we needed $n \mid 90$).
So (D) $90$ is the answer.

$$9, 25, 45, 90 \mid 90 \text{?}\ 9\,\checkmark,\ 25\,\times,\ 45\,\checkmark,\ 90\,\checkmark;\ 120\,\times \;\Rightarrow\; \textbf{(D)}$$

💡 Among the choices, only divisors of $90$ are achievable; biggest one is $90$.

[1] #9 6.NS.C.6 Warm up with a smaller target. How many consecutive integers can sum to $5$? Use

[2] #5 6.NS.C.6 Generalize the trick. For a target sum $S > 0$, take the integers from $-(S - 1)

[3] #5 6.NS.C.6 Apply to $S = 45$. Take the integers from $-44$ up to $45$. That is $-44, -43, \

[4] #9 6.EE.B.7 Confirm this is the maximum. A block of $n$ consecutive integers starting at $a$

[5] #3 4.OA.B.4 Eliminate the smaller choices. (A) $9$, (B) $25$, (C) $45$ are all divisors of $

Review

Reasonableness: Verify directly. The block $-44, -43, \ldots, 44, 45$ has $90$ terms. Pair each negative with its positive: $(-44) + 44 = 0$, $(-43) + 43 = 0$, $\ldots$, $(-1) + 1 = 0$. That's $44$ cancellations leaving $0$ and $45$. The full sum is $0 + 45 = 45\ \checkmark$. So $90$ consecutive integers can indeed sum to $45$ — the answer (D) is achievable.

Alternative: Tool #1 (Diagram): draw a number line and mark the block from $-44$ to $45$. The symmetric portion $[-44, 44]$ around $0$ visibly cancels (mirror image), and only the rightmost dot $45$ contributes to the sum — instantly explaining both why the trick works and why the count is $90$.

CCSS standards used (min grade 6)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Checking which answer choices divide $90$ (and so could be valid block lengths).)
6.NS.C.6 Understand a rational number as a point on the number line (Using positive and negative integers symmetrically around $0$ so most terms cancel in pairs.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Setting $n(2a + n - 1) = 90$ to prove $n$ must divide $90$, hence the maximum is $90$.)

⭐ This AMC 10 problem only needs Grade 6 "negatives on the number line" you already know — start at $-44$ and run up to $45$ so every pair $(-k, k)$ cancels and only $45$ is left, giving $90$ terms.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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