AMC 8 · 2015 · #14
Grade 4 algebranumber-theoryProblem
Which of the following integers cannot be written as the sum of four consecutive odd integers?
Pick an answer.
AMC 8 2015 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: Among the five numbers $16$, $40$, $72$, $100$, and $200$, find the one that can NOT be written as the sum of four consecutive odd integers (like $1+3+5+7$ or $7+9+11+13$).
Givens: We are adding exactly four odd integers in a row; Consecutive odd integers differ by $2$ (e.g., $3, 5, 7, 9$); Answer choices: (A) $16$, (B) $40$, (C) $72$, (D) $100$, (E) $200$
Unknowns: Which of the five answer choices is impossible to express as such a sum
Understand
Restated: Among the five numbers $16$, $40$, $72$, $100$, and $200$, find the one that can NOT be written as the sum of four consecutive odd integers (like $1+3+5+7$ or $7+9+11+13$).
Givens: We are adding exactly four odd integers in a row; Consecutive odd integers differ by $2$ (e.g., $3, 5, 7, 9$); Answer choices: (A) $16$, (B) $40$, (C) $72$, (D) $100$, (E) $200$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
We do not yet know which sums are possible, so start by listing a few real sums of four consecutive odd integers (Tool #9): $1+3+5+7$, $3+5+7+9$, $5+7+9+11$, and so on. Then look at the resulting sums for a pattern (Tool #5) — every sum jumps by $8$, so every possible sum is a multiple of $8$. Finally test the five answer choices against that pattern (Tool #3, Eliminate) — only one choice fails the divisibility-by-$8$ test.
Execute — Answer: D
4.NBT.B.4 Step 1 - Try the easiest cases first.
- Compute the sum of four consecutive odd integers for the smallest starting values and write the answers in order.
💡 Doing several small additions by hand is a Grade 4 multi-digit addition skill, and it gives us real data to look at.
4.OA.C.5 Step 2 - Look at the sequence of sums and find the pattern.
- The sums are $16, 24, 32, 40, \ldots$ Each one is $8$ more than the previous, so every possible sum is a multiple of $8$.
💡 Spotting a constant jump of $8$ in a list of numbers and using that to describe every term is exactly the Grade 4 "generate and analyze patterns" idea. (Why $8$? Each odd integer goes up by $2$, and there are $4$ of them, so the sum goes up by $4 \times 2 = 8$.)
4.OA.B.4 Step 3 - Now test each answer choice: is it a multiple of $8$?
- Whatever is NOT a multiple of $8$ cannot be a sum of four consecutive odd integers.
💡 Checking each choice against the rule "must be a multiple of $8$" is the Grade 4 multiples-and-factors test, and on a multiple-choice problem it is the fastest way to finish.
4.OA.B.4 Step 4 - Only $100$ failed the multiple-of-$8$ test, so $100$ is the one number on the list that cannot be written as the sum of four consecutive odd integers.
- The answer is $\textbf{(D)}$.
💡 Exactly one choice survives the elimination, which matches the problem's promise of one impossible value.
4.NBT.B.4 Try the easiest cases first. Compute the sum of four consecutive odd integers fo 4.OA.C.5 Look at the sequence of sums and find the pattern. The sums are $16, 24, 32, 40, 4.OA.B.4 Now test each answer choice: is it a multiple of $8$? Whatever is NOT a multiple 4.OA.B.4 Only $100$ failed the multiple-of-$8$ test, so $100$ is the one number on the li Review
Reasonableness: Double-check by trying to actually build $100$ as a sum of four consecutive odd integers. The middle of those four integers would be near $100 \div 4 = 25$, so try $23 + 25 + 27 + 29 = 104$ (too big) and $21 + 23 + 25 + 27 = 96$ (too small). The sum jumps straight from $96$ to $104$ — skipping $100$ — which confirms $100$ is impossible. Each of the other four choices can be hit: $16 = 1+3+5+7$, $40 = 7+9+11+13$, $72 = 15+17+19+21$, $200 = 47+49+51+53$.
Alternative: Tool #13 (Convert to Algebra): let the four consecutive odd integers be $n, n+2, n+4, n+6$ where $n$ is odd. Their sum is $4n + 12 = 4(n+3)$. Since $n$ is odd, $n+3$ is even, so $4(n+3)$ is a multiple of $8$. This proves the pattern rigorously, but the small-cases + pattern path is faster and reaches the same conclusion without algebra.
CCSS standards used (min grade 4)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing the first few sums $1+3+5+7=16$, $3+5+7+9=24$, $5+7+9+11=32$, $7+9+11+13=40$ to gather data for a pattern.)4.OA.C.5Generate a number pattern that follows a given rule and identify apparent features (Noticing that the sums $16, 24, 32, 40, \ldots$ increase by $8$ each time and concluding that every such sum is a multiple of $8$.)4.OA.B.4Find all factor pairs and recognize that a whole number is a multiple of each of its factors (Testing each answer choice for divisibility by $8$ to see which one is not a multiple of $8$ — $100 = 8 \times 12 + 4$ fails, all others pass.)
⭐ Adding four small odd numbers a few times and spotting the jump-by-$8$ pattern is a Grade 4 skill — no algebra needed to crack this AMC 8 problem!
⭐ Adding four small odd numbers a few times and spotting the jump-by-$8$ pattern is a Grade 4 skill — no algebra needed to crack this AMC 8 problem!
More like this
Same archetype — closest grade level first.