AMC 8 · 2018 · #19

Easy mode Grade 4

Problem

Picture a pyramid made of boxes. The bottom row has $4$ boxes. The next row up has $3$ boxes, sitting on top of the gaps. Then a row of $2$ , and one box at the very top.

Every box holds either a "+" or a "-".

Here is the rule for the upper rows. Look at any box that is not on the bottom. It sits on top of two boxes from the row below.

If those two boxes below match (both "+" or both "-"), this box gets a "+".
If those two boxes below do not match (one "+" and one "-"), this box gets a "-".

(The diagram below shows one example of the pyramid filled in.)

You get to choose the $4$ signs in the bottom row. Once they are chosen, every other box is forced by the rule.

How many different ways can you fill the bottom row so that the box at the very top ends up as a "+"?

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

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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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