AMC 8 · 2018 · #19

Grade 4 countingpattern

paritypattern-recognitionsystematic-enumerationcombinations-basic pattern-recognitioncaseworktree-enumeration ↑ Prerequisites: paritycombinations-basic

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Problem

In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?

$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: A sign pyramid has $4$ rows. The bottom row has $4$ cells; each cell holds either a $+$ or a $-$. The rule for filling every cell above the bottom is: write $+$ if the two cells directly below it match, write $-$ if they differ. Among all ways to fill the $4$ bottom cells, how many produce a $+$ at the very top?

Givens: Bottom row has $4$ cells, each independently $+$ or $-$; Total number of ways to fill the bottom row $= 2^4 = 16$; Combining rule: same $\to +$, different $\to -$ (this is exactly the XOR-style rule shown in the worked example diagram); Answer choices: (A) $2$, (B) $4$, (C) $8$, (D) $12$, (E) $16$

Unknowns: The number of bottom-row fillings (out of $16$) that make the top cell a $+$

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #9 Solve an Easier Related Problem, #5 Look for a Pattern

There are only $2^4 = 16$ possible bottom rows, which is small enough to list completely (Tool #2). To stay organized and to spot why the answer must be exactly half of $16$, we first solve the same puzzle for a tiny pyramid with a $2$-cell bottom row (Tool #9 Easier Problem) and a $3$-cell bottom row, then look for the pattern (Tool #5). The pattern says: in a sign pyramid, flipping the leftmost bottom cell always flips the top sign, so exactly half of all bottom fillings give a $+$. That same reasoning, verified by listing, gives the answer for $4$ cells without any algebra.

Execute — Answer: C

#2 Make a Systematic List 4.OA.C.5 Step 1

Set up the rule cleanly.
Reading the diagram in the problem, each cell above is decided by the two below: $(+,+)\to+$, $(-,-)\to+$, $(+,-)\to-$, $(-,+)\to-$.
So a cell is $+$ exactly when its two children are the same sign.

$$\text{cell above} = +\text{ if children match},\;-\text{ if children differ}$$

💡 This is a Grade 4 "generate a pattern from a given rule" setup: a clear input-to-output rule we will apply over and over.

#9 Solve an Easier Related Problem 4.OA.C.5 Step 2

Start with the easier problem: a tiny pyramid whose bottom row has only $2$ cells.
There are $2^2 = 4$ bottom rows.
Apply the rule to each: $(+,+)\to+$, $(+,-)\to-$, $(-,+)\to-$, $(-,-)\to+$.
Exactly $2$ of the $4$ rows give a $+$ on top — that is half.

$$\text{top}=+\;\text{count for }n=2\;:\;2\text{ out of }4\;=\;\tfrac{1}{2}$$

💡 Shrinking the problem (Tool #9) lets a Grade 4 student check the rule by hand and see what fraction of fillings win.

#2 Make a Systematic List 4.OA.C.5 Step 3

Try the next case: bottom row of $3$ cells.
There are $2^3 = 8$ bottom rows.
List them in binary order (writing $0$ for $+$ and $1$ for $-$) and apply the rule twice to reach the top: $+\!+\!+\to+$, $+\!+\!-\to-$, $+\!-\!+\to+$, $+\!-\!-\to-$, $-\!+\!+\to-$, $-\!+\!-\to+$, $-\!-\!+\to-$, $-\!-\!-\to+$.
The tops that are $+$ are $4$ out of $8$ — again exactly half.

$$\text{top}=+\;\text{count for }n=3\;:\;4\text{ out of }8\;=\;\tfrac{1}{2}$$

💡 Listing in a fixed order (Tool #2) makes the count reliable; the Grade 4 pattern rule is applied row by row.

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

Look for the pattern (Tool #5).
For $n=2$ and $n=3$, exactly half of all bottom rows produce a $+$ at the top.
The reason is simple to see from any filling: flip just the leftmost cell.
That flip changes the cell directly above it from $+$ to $-$ (or vice versa), which flips the next cell above it, and so on all the way to the top.
So flipping the leftmost bottom cell always flips the top sign — meaning bottom rows pair up perfectly into "top is $+$" and "top is $-$" partners.

$$\#\{\text{top}=+\}\;=\;\#\{\text{top}=-\}\;=\;\tfrac{1}{2}\cdot 2^{n}\;=\;2^{\,n-1}$$

💡 Spotting that the rule is reversible in one spot (Grade 4 pattern reasoning) explains why the count is always exactly half.

#2 Make a Systematic List 4.OA.C.5 Step 5

Apply the pattern to the actual problem ($n=4$ bottom cells).
Total bottom rows $= 2^4 = 16$, and half of them give a $+$ at the top, so the count is $16/2 = 8$.
To double-check, this matches a direct listing: the $8$ "good" bottom rows are $++++$, $+--+$, $-++-$, $----$, $+-+-$, $-+-+$, $++--$, $--++$.

$$\#\{\text{top}=+\}\;=\;\tfrac{1}{2}\cdot 2^{4}\;=\;\tfrac{1}{2}\cdot 16\;=\;8\;\Rightarrow\;\textbf{(C)}$$

💡 A Grade 4 student can both apply the half-rule and verify by listing the $8$ winners explicitly.

[1] #2 4.OA.C.5 Set up the rule cleanly. Reading the diagram in the problem, each cell above is

[2] #9 4.OA.C.5 Start with the easier problem: a tiny pyramid whose bottom row has only $2$ cell

[3] #2 4.OA.C.5 Try the next case: bottom row of $3$ cells. There are $2^3 = 8$ bottom rows. Lis

[4] #5 4.OA.C.5 Look for the pattern (Tool #5). For $n=2$ and $n=3$, exactly half of all bottom

[5] #2 4.OA.C.5 Apply the pattern to the actual problem ($n=4$ bottom cells). Total bottom rows

Review

Reasonableness: The total number of bottom fillings is $2^4 = 16$, so any answer above $16$ would be impossible — choice (E) is the maximum reasonable bound. Both very small ($2$) and very large ($16$) answers would be suspicious because the pyramid rule is symmetric in $+$ and $-$ (swapping every sign in the bottom flips every cell, and the top of an even-height pyramid behaves predictably), suggesting the count should be exactly half. Half of $16$ is $8$, which is choice (C). The explicit list of $8$ bottom rows in the last step confirms the count and matches the answer.

Alternative: Tool #11 (Work Backwards) starts at the desired top $+$ and asks what Row $3$ can be: $++$ or $--$. For each of those, walk back to Row $2$, then to the bottom. Casework gives $4+4=8$ valid bottom rows. This route also uses only Grade 4 "apply a rule" thinking — no algebra needed — and lands on the same answer (C).

CCSS standards used (min grade 4)

4.OA.C.5 Generate a number or shape pattern following a given rule (Applying the sign-pyramid rule (same $\to +$, different $\to -$) row by row for the $n=2$, $n=3$, and $n=4$ cases, listing bottom rows systematically, and recognizing the "exactly half are $+$" pattern that gives the final count of $8$.)

⭐ This AMC 8 problem only needs Grade 4 "follow a rule and look for the pattern" thinking you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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