AMC 8 · 2024 · #5

Grade 4 number-theory

Archetype Small-Domain Constrained Search

divisibility-rulesfactorsmultiples caseworksystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticfactors

📏 Long solution 💡 3 insights

Problem

Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of $6$ . Which of the following integers cannot be the sum of the two numbers?

$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Aaliyah rolls two standard 6-sided dice (each die shows a number from 1 to 6). The product of the two numbers happens to be a multiple of 6. Among the listed integers 5, 6, 7, 8, 9, we must find the one that CANNOT possibly be the sum of those two dice.

Givens: Two standard dice, so each value $d_1, d_2$ is an integer with $1 \le d_1, d_2 \le 6$; The product $d_1 \times d_2$ is a multiple of $6$; Five candidate sums to test: (A) 5, (B) 6, (C) 7, (D) 8, (E) 9

Unknowns: Which candidate sum is impossible — i.e. cannot be written as $d_1 + d_2$ for any pair whose product is divisible by 6

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #3 Eliminate Possibilities, #7 Identify Subproblems

There are only 5 candidate sums and at most a handful of dice pairs for each, so we can simply LIST every pair $(d_1, d_2)$ with $1 \le d_1 \le d_2 \le 6$ that adds to a given sum (Tool #2). For each sum we ELIMINATE the candidate the moment we find one pair whose product is a multiple of 6 — that proves the sum is reachable (Tool #3). The whole task splits into the SUBPROBLEM 'is $d_1 d_2$ divisible by 6?', which we break further into 'is it divisible by 2?' and 'is it divisible by 3?' (Tool #7). No algebra is needed.

Execute — Answer: B

#7 Identify Subproblems 4.OA.B.4 Step 1

First, restate what 'product is a multiple of 6' really demands.
Since $6 = 2 \times 3$, the product $d_1 \times d_2$ is a multiple of 6 exactly when it is divisible by 2 AND divisible by 3.
That means among the two dice we need at least one even number (2, 4, or 6) and at least one multiple of 3 (3 or 6).
This is the divisibility subproblem; once we have this rule we can scan pairs very quickly.

$$6 = 2 \times 3 \Rightarrow d_1 \times d_2 \equiv 0 \pmod{6} \iff (\text{at least one even}) \text{ AND } (\text{at least one multiple of }3)$$

💡 Grade 4 students learn that 'multiple of 6' means a number you can build with 2s and 3s, so we split the divisibility test into two easy checks.

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

Test sum = 5 by listing every pair $(d_1, d_2)$ with $d_1 \le d_2$ that adds to 5, then computing the product.
Pair $(1,4)$ gives product 4 — not a multiple of 6.
Pair $(2,3)$ gives product 6 — a multiple of 6!
So sum 5 IS achievable, and we eliminate choice (A).

$1+4=5,\ 1\times 4 = 4 \;\text{(no)}$ ; $\ 2+3=5,\ 2\times 3 = 6 \;\checkmark$

💡 Listing every pair that adds to a single-digit number uses Grade 2 fact fluency within 20.

#2 Make a Systematic List 3.OA.C.7 Step 3

Test sum = 6.
The pairs are $(1,5), (2,4), (3,3)$ with products $5$, $8$, and $9$.
None of these products is a multiple of 6 (we need both a factor of 2 and a factor of 3, but each pair is missing one of them).
So NO pair adding to 6 satisfies the condition — sum 6 looks impossible.

$1\times 5 = 5\ (\text{no})$ ; $\ 2\times 4 = 8\ (\text{no})$ ; $\ 3\times 3 = 9\ (\text{no})$

💡 Multiplying single-digit factors and checking divisibility uses Grade 3 multiplication fluency within 100.

#2 Make a Systematic List 3.OA.C.7 Step 4

Test sum = 7.
Pairs $(1,6), (2,5), (3,4)$ give products 6, 10, 12.
Both 6 and 12 are multiples of 6, so sum 7 IS achievable — eliminate choice (C).

$1\times 6 = 6\ \checkmark$ ; $\ 2\times 5 = 10\ (\text{no})$ ; $\ 3\times 4 = 12\ \checkmark$

💡 Once you know $6, 12, 18, \ldots$ are multiples of 6, the check is instant Grade 3 multiplication.

#2 Make a Systematic List 3.OA.C.7 Step 5

Test sum = 8.
Pairs $(2,6), (3,5), (4,4)$ give products 12, 15, 16.
The pair $(2,6)$ gives 12, a multiple of 6, so sum 8 IS achievable — eliminate choice (D).

$2\times 6 = 12\ \checkmark$ ; $\ 3\times 5 = 15\ (\text{no})$ ; $\ 4\times 4 = 16\ (\text{no})$

💡 Listing pairs and multiplying is still Grade 3 work — no algebra needed.

#2 Make a Systematic List 3.OA.C.7 Step 6

Test sum = 9.
Pairs $(3,6), (4,5)$ give products 18 and 20.
The pair $(3,6)$ gives 18, a multiple of 6, so sum 9 IS achievable — eliminate choice (E).

$3\times 6 = 18\ \checkmark$ ; $\ 4\times 5 = 20\ (\text{no})$

💡 $3 \times 6 = 18 = 6 \times 3$ uses the same Grade 3 multiplication facts.

#3 Eliminate Possibilities 3.OA.D.8 Step 7

Four of the five choices (A, C, D, E) have been eliminated because we exhibited an explicit pair whose product is a multiple of 6.
Only choice (B) sum = 6 survives — no pair adding to 6 has a product divisible by 6.
So 6 is the integer that CANNOT be the sum.

$$\text{Eliminated: } 5, 7, 8, 9 \Rightarrow \text{Impossible sum} = 6 \Rightarrow \textbf{(B)}$$

💡 Putting the casework together to pick the one remaining answer is Grade 3 multi-step problem solving.

[1] #7 4.OA.B.4 First, restate what 'product is a multiple of 6' really demands. Since $6 = 2 \t

[2] #2 2.OA.B.2 Test sum = 5 by listing every pair $(d_1, d_2)$ with $d_1 \le d_2$ that adds to

[3] #2 3.OA.C.7 Test sum = 6. The pairs are $(1,5), (2,4), (3,3)$ with products $5$, $8$, and $9

[4] #2 3.OA.C.7 Test sum = 7. Pairs $(1,6), (2,5), (3,4)$ give products 6, 10, 12. Both 6 and 12

[5] #2 3.OA.C.7 Test sum = 8. Pairs $(2,6), (3,5), (4,4)$ give products 12, 15, 16. The pair $(2

[6] #2 3.OA.C.7 Test sum = 9. Pairs $(3,6), (4,5)$ give products 18 and 20. The pair $(3,6)$ giv

[7] #3 3.OA.D.8 Four of the five choices (A, C, D, E) have been eliminated because we exhibited

Review

Reasonableness: Why is sum 6 the special case? A multiple of 6 needs BOTH a factor of 2 AND a factor of 3. The dice values that supply a factor of 3 are only 3 and 6. If one die is 6, the other must be $6 - 6 = 0$ (impossible). If one die is 3, the other must be $6 - 3 = 3$, but then the product is $3 \times 3 = 9$, which has no factor of 2. So no pair adding to 6 can carry both factors — exactly matching what our casework showed. The answer (B) 6 is consistent with the structural reason.

Alternative: An alternative is Tool #16 (Change Focus / Complement): instead of testing each sum, first list every dice pair whose product IS a multiple of 6 — namely $(1,6), (2,3), (2,6), (3,4), (3,6), (4,6), (5,6), (6,6)$ — then compute their sums $\{7, 5, 8, 7, 9, 10, 11, 12\}$. The set of achievable sums is $\{5, 7, 8, 9, 10, 11, 12\}$. The missing sum from the answer list is 6, confirming (B).

CCSS standards used (min grade 4)

2.OA.B.2 Fluently add and subtract within 20 using mental strategies (Listing all pairs of dice values that add to each candidate sum 5-9.)
3.OA.C.7 Fluently multiply and divide within 100 (Computing each pair's product (e.g. $2 \times 3 = 6$, $4 \times 4 = 16$) so we can test divisibility by 6.)
3.OA.D.8 Solve two-step word problems using four operations within 100 (Combining the case-by-case results to select the single sum that cannot be produced.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Recognizing 'multiple of 6' and splitting it into the joint conditions 'divisible by 2 AND divisible by 3'.)

⭐ This AMC 8 problem only needs Grade 4 multiples-and-factors thinking you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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