AMC 8 · 2016 · #19

Easy mode Grade 6

Problem

Picture a list of $25$ even numbers in a row, where each one is $2$ more than the one before. For example, a list like $2, 4, 6, 8, \ldots$ would follow this pattern, but the list here starts at a different number.

When you add up all $25$ of these numbers, the total is $10{,}000$ .

What is the largest number in the list?

Pick an answer.

(A)

$mbox{ }360$

(B)

$mbox{ }388$

(C)

$mbox{ }412$

(D)

$mbox{ }416$

(E)

$mbox{ }424$

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Toolkit + CCSS Solution

Understand

Restated: Take $25$ even integers in a row, like $..., n-4, n-2, n, n+2, n+4, ...$. Their sum is $10{,}000$. What is the largest one?

Givens: There are exactly $25$ integers; They are consecutive even integers (each is $2$ more than the previous); Their total sum is $10{,}000$; Answer choices: (A) $360$, (B) $388$, (C) $412$, (D) $416$, (E) $424$

Unknowns: The largest of the $25$ integers (the $25^{\text{th}}$ term)

Understand

Restated: Take $25$ even integers in a row, like $..., n-4, n-2, n, n+2, n+4, ...$. Their sum is $10{,}000$. What is the largest one?

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems

Twenty-five consecutive even integers is too many to list, but the structure is highly regular. Tool #9 (Easier Problem) — try $3$ or $5$ consecutive evens first — exposes the pattern that the sum is always (count) $\times$ (middle term). Tool #5 (Pattern) lets us use that fact to jump straight to the middle term: $10{,}000 \div 25 = 400$. Then Tool #7 (Subproblems) splits the rest into two clean pieces — find the middle term, then step out $12$ places (by $+2$ each) to reach the largest. This dodges Tool #13 (Algebra) entirely.

Execute — Answer: E

#9 Solve an Easier Related Problem 3.OA.B.5 Step 1

Try the easier version with $3$ consecutive evens.
Take $4, 6, 8$: sum is $18$ and the middle term is $6$.
Notice $18 = 3 \times 6$.
Try $5$ consecutive evens $2, 4, 6, 8, 10$: sum is $30 = 5 \times 6$, and the middle term is $6$ again.
The pattern: for an odd number of evenly-spaced terms, sum $=$ count $\times$ middle term.

$4 + 6 + 8 = 18 = 3 \times 6$, \quad $2 + 4 + 6 + 8 + 10 = 30 = 5 \times 6$

💡 Working a small case before tackling $25$ terms shows why pairs above and below the middle cancel — a Grade 3 "properties of operations" idea.

#5 Look for a Pattern 6.SP.B.5 Step 2

Apply the pattern to the real problem.
The $25$ even integers have an odd count, so their sum equals $25 \times$ (middle term).
Divide both sides by $25$ to recover the middle term.

$$\text{middle term} = \dfrac{10{,}000}{25} = 400$$

💡 For an odd-length evenly-spaced list, the mean and the middle value are the same number — a Grade 6 statistics fact.

#7 Identify Subproblems 4.OA.C.5 Step 3

Locate the middle position.
With $25$ terms, the middle is the $13^{\text{th}}$ term (because $12$ sit to its left and $12$ to its right).
So term $13$ equals $400$.

$$\dfrac{25 + 1}{2} = 13$$

💡 Splitting the count into "left of middle", "middle", "right of middle" is a Grade 4 sequence-position move.

#5 Look for a Pattern 4.OA.C.5 Step 4

Step from the $13^{\text{th}}$ term up to the $25^{\text{th}}$ term.
That's $25 - 13 = 12$ steps, and each step adds $2$ because the integers are consecutive evens.

$$\text{largest} = 400 + 12 \times 2 = 400 + 24 = 424 \;\Rightarrow\; \textbf{(E)}$$

💡 Adding the common difference $12$ times to walk from term $13$ to term $25$ is the same skip-counting Grade 4 students do with patterns.

[1] #9 3.OA.B.5 Try the easier version with $3$ consecutive evens. Take $4, 6, 8$: sum is $18$ a

[2] #5 6.SP.B.5 Apply the pattern to the real problem. The $25$ even integers have an odd count,

[3] #7 4.OA.C.5 Locate the middle position. With $25$ terms, the middle is the $13^{\text{th}}$

[4] #5 4.OA.C.5 Step from the $13^{\text{th}}$ term up to the $25^{\text{th}}$ term. That's $25

Review

Reasonableness: Sanity check the smallest term too: $400 - 12 \times 2 = 376$. So the list runs $376, 378, \ldots, 424$. Quick pairing: $(376 + 424) = 800$, and pairing every term with its mirror around the middle gives $12$ pairs of $800$ plus the lone middle $400$, total $12 \times 800 + 400 = 9600 + 400 = 10{,}000$. \checkmark\ The answer $424$ is the largest choice given and is consistent with the middle being a round $400$.

Alternative: Tool #3 (Eliminate Possibilities) on the choices: if the largest is $L$, the smallest is $L - 48$, and the sum is $25 \times \dfrac{L + (L-48)}{2} = 25(L - 24)$. Set $25(L - 24) = 10{,}000$ so $L - 24 = 400$ and $L = 424$. Only choice (E) works; (A)-(D) each fail this equation.

CCSS standards used (min grade 6)

3.OA.B.5 Apply properties of operations as strategies to multiply and divide (Recognizing on the small cases ($3$ and $5$ terms) that the sum factors as count $\times$ middle term — a use of the distributive/commutative properties.)
4.OA.C.5 Generate a number or shape pattern that follows a given rule (Locating the $13^{\text{th}}$ term as the middle of $25$ and then stepping $12$ times by $+2$ to reach the $25^{\text{th}}$ term.)
6.SP.B.5 Summarize numerical data sets in relation to their context (mean, median) (Using the fact that for an evenly-spaced list with an odd number of terms, the mean ($10{,}000 / 25 = 400$) equals the middle value.)

⭐ If you have an odd number of evenly-spaced numbers, the average IS the middle one — a Grade 6 idea that cracks this AMC 8 problem in one division!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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