AMC 10 · 2019 · #5
Grade 6 arithmeticProblem
What is the greatest number of consecutive integers whose sum is 45?
Pick an answer.
AMC 10 2019 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.
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 $45$.
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) $120$
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
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 $\{2, 3\}$.
💡 Negatives cancel positives in pairs — the only 'survivor' is the top number.
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.
💡 Pattern from the warm-up — sum survives, length doubles.
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 $45 - (-44) + 1 = 90$.
💡 Same pattern as the warm-up, just bigger: from $-(S-1)$ to $S$ gives $2S$ terms.
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.
💡 Length must divide $90$ — and $90$ itself works.
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.
💡 Among the choices, only divisors of $90$ are achievable; biggest one is $90$.
6.NS.C.6 Warm up with a smaller target. How many consecutive integers can sum to $5$? Use 6.NS.C.6 Generalize the trick. For a target sum $S > 0$, take the integers from $-(S - 1) 6.NS.C.6 Apply to $S = 45$. Take the integers from $-44$ up to $45$. That is $-44, -43, \ 6.EE.B.7 Confirm this is the maximum. A block of $n$ consecutive integers starting at $a$ 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.4Find 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.6Understand 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.7Solve 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.
⭐ 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.
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