AMC 8 · 2016 · #8
Easy mode Grade 5Problem
Look at this long expression:
100−98+96−94+92−90+⋯+8−6+4−2.
The numbers used are every even number from 2 up to 100. The signs alternate: the bigger one in each pair has a plus sign, the smaller one has a minus sign.
If you pair them up two at a time, each pair becomes a small piece you can calculate easily.
What is the value of the whole expression?
Pick an answer.
AMC 8 2016 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: Compute the value of the alternating sum $100-98+96-94+92-90+\cdots+8-6+4-2$, where the terms are consecutive even numbers from $100$ down to $2$ with signs alternating $+,-,+,-,\dots$ starting from a positive $100$.
Givens: The expression lists every even integer from $100$ down to $2$; Signs strictly alternate: $+100, -98, +96, -94, \dots, +4, -2$; Consecutive even numbers differ by $2$; Answer choices: (A) $20$, (B) $40$, (C) $50$, (D) $80$, (E) $100$
Unknowns: The numerical value of the entire alternating sum
Understand
Restated: Compute the value of the alternating sum $100-98+96-94+92-90+\cdots+8-6+4-2$, where the terms are consecutive even numbers from $100$ down to $2$ with signs alternating $+,-,+,-,\dots$ starting from a positive $100$.
Givens: The expression lists every even integer from $100$ down to $2$; Signs strictly alternate: $+100, -98, +96, -94, \dots, +4, -2$; Consecutive even numbers differ by $2$; Answer choices: (A) $20$, (B) $40$, (C) $50$, (D) $80$, (E) $100$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
Computing $50$ signed terms one at a time is slow and error-prone. Tool #7 (Identify Subproblems) lets us regroup the long sum into $(100-98)+(96-94)+\cdots+(4-2)$, turning one giant problem into many tiny copies of the same subproblem. Tool #5 (Look for a Pattern) then spots that every single pair evaluates to $2$, so the original sum collapses to $2+2+\cdots+2$. Tool #9 (Solve an Easier Related Problem) is held in reserve as a sanity check — we can verify the method on the much shorter sum $4-2$ or $8-6+4-2$ before trusting it on the full $50$-term version.
Execute — Answer: C
5.OA.A.1 Step 1 - Use the associative property of addition to group consecutive terms into pairs.
- Because the signs go $+,-,+,-,\dots$, each pair has a positive even number followed by the next-smaller even number with a minus sign.
💡 Inserting parentheses to group $+a-b$ pairs is exactly the Grade 5 "use parentheses in expressions" move — same value, easier shape.
4.OA.C.5 Step 2 - Evaluate the first few pairs to see if a pattern shows up.
- Each pair is a positive even number minus the even number that is $2$ smaller, so each subtraction gives $2$.
💡 Spotting that every pair equals the same constant ($2$) is the Grade 4 "analyze a generated pattern" skill in action.
4.OA.C.5 Step 3 - Count how many such pairs there are.
- The expression lists every even number from $2$ to $100$, which is $\tfrac{100}{2}=50$ numbers.
- Pairs use $2$ numbers each, so the number of pairs is $\tfrac{50}{2}=25$.
💡 Even numbers from $2$ to $100$ correspond one-to-one with $1, 2, 3, \dots, 50$ — a Grade 4 pattern-counting argument.
3.OA.A.1 Step 4 - The whole sum is now $25$ copies of $2$ added together.
- That is the very definition of multiplication: $25$ equal groups of $2$.
💡 Turning a repeated sum into a multiplication is the Grade 3 "products as equal groups" definition.
5.OA.A.1 Use the associative property of addition to group consecutive terms into pairs. 4.OA.C.5 Evaluate the first few pairs to see if a pattern shows up. Each pair is a positi 4.OA.C.5 Count how many such pairs there are. The expression lists every even number from 3.OA.A.1 The whole sum is now $25$ copies of $2$ added together. That is the very definit Review
Reasonableness: A rough estimate confirms the size. The sum starts with $+100$ but is immediately knocked down by $-98$, $+96$ by $-94$, and so on — every positive is canceled to within $2$ by the next term. So the running total grows by only about $2$ per pair, not by hundreds. With $25$ pairs, we expect something near $25 \times 2 = 50$ — exactly choice (C). Choices like $80$ or $100$ are far too big given how aggressively the negatives cancel the positives.
Alternative: Tool #9 (Solve an Easier Related Problem): try the shorter sum $8-6+4-2$. By hand: $8-6=2$, $4-2=2$, total $=4$. There are $4$ terms (the even numbers $2,4,6,8$), making $2$ pairs of value $2$, and $2 \times 2 = 4$ matches. The same recipe applied to $50$ terms gives $25 \times 2 = 50$, confirming (C).
CCSS standards used (min grade 5)
5.OA.A.1Use parentheses, brackets, or braces in numerical expressions (Regrouping $100-98+96-94+\cdots+4-2$ as $(100-98)+(96-94)+\cdots+(4-2)$ using the associative property — the Grade 5 "parentheses in expressions" standard.)4.OA.C.5Generate and analyze a number or shape pattern (Observing that each pair $(2k)-(2k-2)=2$ produces the same constant value, and counting the $50$ even-number terms via the $1$-to-$1$ correspondence with $1,2,\dots,50$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Performing each two-digit subtraction such as $100-98=2$ and $96-94=2$ to verify the pair value.)3.OA.A.1Interpret products of whole numbers as equal groups (Replacing the repeated sum $\underbrace{2+2+\cdots+2}_{25}$ with the multiplication $25 \times 2 = 50$.)
⭐ Group the $50$ numbers into $25$ pairs; each pair quietly equals $2$, so the whole sum is just $25 \times 2 = 50$.
⭐ Group the $50$ numbers into $25$ pairs; each pair quietly equals $2$, so the whole sum is just $25 \times 2 = 50$.
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