Sensim Math Original · sm-6

SM Original Grade 4 number-theoryarithmetic

Inspired by AMC 8 2024 #4

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

perfect-squaresfactorsprime-factorizationsystematic-enumeration systematic-enumerationcaseworkdigit-constraints ↑ Prerequisites: multi-digit-arithmeticfactorsperfect-squares

📏 Medium solution 💡 3 insights

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Problem

A woodworker has nine round discs, each carved with one of the integers $2, 3, 4, 5, 6, 7, 8, 9, 10$ . She wants to arrange the discs on a display so that the product of every label shown is a perfect square. To make this possible she must set exactly one disc aside; the remaining eight discs go on the display. Which integer is on the disc she sets aside?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A woodworker has nine discs labeled with the integers $2$ through $10$. She must remove exactly one disc; the product of the labels on the remaining eight discs must be a perfect square. Among the answer choices $\{4, 5, 6, 7, 8\}$, which integer is on the removed disc?

Givens: Nine discs are labeled with the integers $2, 3, 4, 5, 6, 7, 8, 9, 10$; Exactly one disc is set aside; the other eight stay on the display; The product of the eight remaining labels must be a perfect square; Five candidate removed labels: (A) 4, (B) 5, (C) 6, (D) 7, (E) 8

Unknowns: The single integer $r \in \{2, 3, \ldots, 10\}$ that is removed so that the product of the other eight numbers is a perfect square

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

There are only five candidates, so we can test each one (Tool #6). To make each test fast we first organize the full product into prime-factor building blocks $2$, $3$, $5$, $7$ (Tool #2) — once we know how many of each prime appear in the total, removing a disc just subtracts that disc's contribution. A prime with an odd total exponent must be brought down to an even exponent by the removed disc, which immediately eliminates any choice that does not carry the right primes (Tool #3). No algebra (Tool #13) is needed.

Execute — Answer: D

#2 Make a Systematic List 4.OA.B.4 Step 1

Write each label $2$ through $10$ as a product of small primes.
This is the "organize" move: instead of staring at the giant product $2 \cdot 3 \cdot 4 \cdots 10$, we list how many copies of each prime $2$, $3$, $5$, $7$ live inside each disc.
The list is short — every label uses only primes up to $7$.

$$2{=}2,\ 3{=}3,\ 4{=}2^2,\ 5{=}5,\ 6{=}2\cdot 3,\ 7{=}7,\ 8{=}2^3,\ 9{=}3^2,\ 10{=}2\cdot 5$$

💡 Grade 4 students learn to break a whole number into its prime factor pairs, so each label becomes a small bundle of $2$s, $3$s, $5$s, and $7$s.

#2 Make a Systematic List 4.OA.B.4 Step 2

Add up how many copies of each prime appear across all nine discs.
Count the $2$s: discs $2, 4, 6, 8, 10$ contribute $1 + 2 + 1 + 3 + 1 = 8$ twos.
Count the $3$s: discs $3, 6, 9$ contribute $1 + 1 + 2 = 4$ threes.
Count the $5$s: discs $5$ and $10$ contribute $1 + 1 = 2$ fives.
Count the $7$s: only disc $7$ contributes $1$ seven.
So the full product is $2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7^{1}$.

$$2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9\cdot 10 = 2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7$$

💡 Counting how many $2$s, $3$s, $5$s, and $7$s come from each disc is Grade 4 factor-and-multiple work.

#6 Guess and Check 3.OA.C.7 Step 3

A whole number is a perfect square exactly when every prime in its factorization has an **even** exponent.
Look at the total: the exponents of $2$, $3$, and $5$ are $8, 4, 2$ — all even — but the exponent of $7$ is $1$, which is odd.
The only way to make every exponent even by removing one disc is to remove the disc that carries the lone factor of $7$.

$\text{exponents}=(8,4,2,1)$ for primes $(2,3,5,7) \;\Rightarrow\; \text{only}\ 7\ \text{is odd}$

💡 Recognizing that a perfect square has even-power primes is the kid-friendly version of $n = k^2$, building on Grade 3 multiplication fluency.

#3 Eliminate Possibilities 4.OA.B.4 Step 4

Check the five candidates by reading off their prime factors.
(A) $4 = 2^2$ — removing it leaves $7$ with odd exponent; not a square.
(B) $5$ — removing it makes the $5$-exponent odd; not a square.
(C) $6 = 2 \cdot 3$ — removing it makes both the $2$- and $3$-exponents odd; not a square.
(D) $7$ — removing it leaves $2^{8} \cdot 3^{4} \cdot 5^{2}$, all even exponents; this IS a perfect square.
(E) $8 = 2^3$ — removing it leaves the $2$-exponent at $5$ and the $7$-exponent at $1$, both odd; not a square.
Only (D) works.

$$\dfrac{2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7}{7} = 2^{8} \cdot 3^{4} \cdot 5^{2} = (2^{4}\cdot 3^{2}\cdot 5)^{2} = 720^{2}$$

💡 Walking down the answer list and crossing off the four that leave an odd-power prime is Grade 4 factor-and-multiple reasoning.

[1] #2 4.OA.B.4 Write each label $2$ through $10$ as a product of small primes. This is the "org

[2] #2 4.OA.B.4 Add up how many copies of each prime appear across all nine discs. Count the $2$

[3] #6 3.OA.C.7 A whole number is a perfect square exactly when every prime in its factorization

[4] #3 4.OA.B.4 Check the five candidates by reading off their prime factors. (A) $4 = 2^2$ — re

Review

Reasonableness: The removed disc has to absorb every prime that appears an odd number of times in the full product. The only odd-exponent prime is $7$, and the only disc containing a $7$ is the disc labeled $7$ itself, so $7$ is the unique answer — it could not have been any of the other four candidates. Sanity check the resulting product: $2^{8} \cdot 3^{4} \cdot 5^{2} = 256 \cdot 81 \cdot 25 = 518{,}400 = 720 \times 720$. Indeed $518{,}400$ is a perfect square (with side $720$), matching the prime-factor argument.

Alternative: Tool #16 (Change Focus / Complement) gives the same answer faster: instead of testing each disc, just ask "which primes appear an odd number of times in $2 \cdot 3 \cdots 10$?" The only odd-exponent prime is $7$ (because $7$ shows up in exactly one disc, namely the disc $7$). The removed disc must therefore carry a $7$ — and only the disc labeled $7$ qualifies. Answer (D) drops out without any candidate-by-candidate testing.

CCSS standards used (min grade 4)

3.OA.C.7 Fluently multiply and divide within 100 (Recognizing a perfect square as $k \times k$ and reading exponents (e.g. $2 \times 2 = 2^2$) for the small primes used to factor each disc.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Breaking each disc label into its prime factors and counting how many copies of each prime appear in the full product to decide which removal makes every exponent even.)

⭐ This AMC 8 problem only needs the Grade 4 prime-factor counting you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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