Sensim Math Original · sm-9

Easy mode Grade 4

📗 View original problem →

Problem

Picture Mina building a tower out of identical wooden blocks. To start, she sets down $3$ blocks.

After that, Mina does $3$ more rounds. In each round, she picks exactly one of two moves:

Move A: add $2$ more blocks to the top of the tower.
Move B: a helper looks at the tower and copies it, so the number of blocks doubles.

Different choices over the $3$ rounds can lead to different final towers.

How many different tower heights (counted in blocks) are possible after the $3$ rounds?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Mina begins with a tower of $3$ blocks. In each of the next $3$ rounds she independently chooses one of two moves: add $2$ blocks ($+2$) or double the current count ($\times 2$). We need to count how many distinct final tower heights are possible after exactly $3$ rounds.

Givens: Initial height: $3$ blocks; Two possible moves per round: $+2$ or $\times 2$; Exactly $3$ rounds are played, each move chosen independently; Five answer choices: (A) 5, (B) 6, (C) 7, (D) 8, (E) 9

Unknowns: The number of distinct tower heights (in blocks) that can occur after $3$ rounds

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

Since each round has only $2$ choices and there are exactly $3$ rounds, the entire possibility space has at most $2^3 = 8$ paths — small enough to write down every outcome explicitly (Tool #2). Picturing the rounds as a level-by-level branching tree (Tool #1) keeps duplicate heights from sneaking by: when two different paths land on the same number, the tree shows it visibly. Once the leaf list is in hand, deduplicating and matching against the (A)-(E) choices is a clean Tool #3 step. Algebra is unnecessary here.

Execute — Answer: C

#2 Make a Systematic List 2.OA.B.2 Step 1

Apply the two possible moves to the starting tower of $3$ blocks to enumerate every height after round $1$.
The two children of the root are $3 + 2 = 5$ (add two) and $3 \times 2 = 6$ (double).
So after round $1$ the height is in $\{5, 6\}$.

$$3 + 2 = 5,\quad 3 \times 2 = 6 \;\Rightarrow\; \text{Round 1} \in \{5,\,6\}$$

💡 Single-digit addition and doubling within $20$ are fluent Grade 2 mental-math operations.

#1 Draw a Diagram 3.OA.C.7 Step 2

Branch each of round $1$'s two heights again to find every round-$2$ outcome.
From $5$: $5 + 2 = 7$ and $5 \times 2 = 10$.
From $6$: $6 + 2 = 8$ and $6 \times 2 = 12$.
No two of the four results coincide, so after round $2$ the distinct heights are $\{7, 8, 10, 12\}$.

$$\begin{aligned} 5 &\to 5+2=7,\ 5\times 2=10 \\ 6 &\to 6+2=8,\ 6\times 2=12 \end{aligned} \;\Rightarrow\; \text{Round 2} \in \{7,\,8,\,10,\,12\}$$

💡 Drawing the branching tree level by level makes it easy to spot any collision; doublings like $5\times 2$ and $6\times 2$ are Grade 3 multiplication fluency.

#2 Make a Systematic List 3.OA.D.8 Step 3

Apply the two moves one more time to each of the four round-$2$ heights to produce every possible round-$3$ height.
From $7$: $9, 14$.
From $8$: $10, 16$.
From $10$: $12, 20$.
From $12$: $14, 24$.
Listing the eight raw outputs gives $\{9, 14, 10, 16, 12, 20, 14, 24\}$.
Notice that $14$ appears twice — both $7 \times 2 = 14$ and $12 + 2 = 14$ land on the same height — so two different paths collide.

$$\begin{aligned} 7 &\to 7+2=9,\ 7\times 2=14 \\ 8 &\to 8+2=10,\ 8\times 2=16 \\ 10 &\to 10+2=12,\ 10\times 2=20 \\ 12 &\to 12+2=14,\ 12\times 2=24 \end{aligned}$$

💡 Carrying the two operations within $100$ across each round-$2$ height is the multi-step, four-operation reasoning of Grade 3.

#3 Eliminate Possibilities 4.OA.C.5 Step 4

Sort the eight raw round-$3$ outputs and remove duplicates: $9, 10, 12, 14, 14, 16, 20, 24$.
The repeated $14$ collapses into one entry, leaving the distinct set $\{9, 10, 12, 14, 16, 20, 24\}$ — a total of $7$ different heights.
This matches answer choice (C).
The other choices are eliminated: (A) $5$ and (B) $6$ are too small (they don't even reach $8$ leaves), (D) $8$ would mean forgetting the $14$ collision, and (E) $9$ overshoots the $2^3 = 8$ leaf maximum.

$$\{9,\,10,\,12,\,14,\,16,\,20,\,24\} \;\Rightarrow\; \#\text{distinct} = 7 \;\Rightarrow\; \textbf{(C)}$$

💡 Generating the terms of a sequence by repeatedly applying a given rule ('$+2$ or $\times 2$') and counting the distinct outputs is exactly the Grade 4 'generate a pattern from a rule' standard.

[1] #2 2.OA.B.2 Apply the two possible moves to the starting tower of $3$ blocks to enumerate ev

[2] #1 3.OA.C.7 Branch each of round $1$'s two heights again to find every round-$2$ outcome. Fr

[3] #2 3.OA.D.8 Apply the two moves one more time to each of the four round-$2$ heights to produ

[4] #3 4.OA.C.5 Sort the eight raw round-$3$ outputs and remove duplicates: $9, 10, 12, 14, 14,

Review

Reasonableness: There are $2 \times 2 \times 2 = 8$ possible move sequences. If no two sequences ever produced the same height, the answer would be $8$. We found exactly one collision at the final level ($7 \times 2 = 14$ and $12 + 2 = 14$), which drops the count from $8$ to $8 - 1 = 7$. So $7$ is consistent with the total path count, and the answer (C) is correct. Every height in $\{9, 10, 12, 14, 16, 20, 24\}$ is between $9$ and $24$, which is a sensible range for $3$ rounds of growth starting from $3$.

Alternative: Tool #1 (Draw a Diagram) standing alone also works: draw a binary tree with $3$ at the root and at every node send a left branch ($+2$) and a right branch ($\times 2$). After three levels there are $8$ leaves; circle each pair of leaves that show the same number and count the colored circles plus the singletons. The tree picture is just a visual rendering of the systematic list above, so either tool delivers the same answer.

CCSS standards used (min grade 4)

2.OA.B.2 Fluently add and subtract within 20 using mental strategies (Computing the round-$1$ transitions $3 + 2 = 5$ and the small doubling $3 \times 2 = 6$ from the starting tower.)
3.OA.C.7 Fluently multiply and divide within 100 (Doubling round-$1$ and round-$2$ heights ($5 \times 2, 6 \times 2, 7 \times 2, 8 \times 2, 10 \times 2, 12 \times 2$) using single-digit multiplication fluency.)
3.OA.D.8 Solve two-step word problems using four operations within 100 (Applying the two moves again to each round-$2$ height to build every round-$3$ candidate in a multi-step calculation.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Repeatedly applying the rule '$+2$ or $\times 2$' for three rounds to generate every possible tower height and counting distinct results.)

⭐ This AMC-style problem only needs the Grade 4 skill of generating number patterns from a rule that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

More like this