AMC 10 · 2024 · #8

Grade 8 arithmetic

divisor-countfactorsunits-digit-trackingprime-factorization identify-subproblemsunits-digit-trackingmodular-arithmetic-mod-10 ↑ Prerequisites: factorsprime-factorizationunits-digit-tracking

📏 Short solution 💡 2 insights

Problem

Let $N$ be the product of all the positive integer divisors of $42$ . What is the units digit
of $N$ ?

$\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Multiply every positive integer that divides $42$ together. Call this product $N$. Find the units digit (ones digit) of $N$.

Givens: $N = $ product of all positive divisors of $42$; $42 = 2 \times 3 \times 7$, so the divisors are $1, 2, 3, 6, 7, 14, 21, 42$; Answer choices: (A) $0$, (B) $2$, (C) $4$, (D) $6$, (E) $8$

Unknowns: The ones digit of the (very large) number $N$

Understand

Restated: Multiply every positive integer that divides $42$ together. Call this product $N$. Find the units digit (ones digit) of $N$.

Givens: $N = $ product of all positive divisors of $42$; $42 = 2 \times 3 \times 7$, so the divisors are $1, 2, 3, 6, 7, 14, 21, 42$; Answer choices: (A) $0$, (B) $2$, (C) $4$, (D) $6$, (E) $8$

Plan

Primary tool: #5 Look for a Pattern

Secondary: #2 Make a Systematic List, #7 Identify Subproblems

The pattern Tool #5 spots is the symmetry of divisors: every divisor $d$ has a unique "partner" $42/d$, and the two multiply to $42$. So once we list the divisors systematically (Tool #2), we can pair them into $42 \times 42 \times 42 \times 42 = 42^4$. Tool #7 (Identify Subproblems) then reduces the units-digit question to just "what is the ones digit of $42^4$?" — which depends only on the ones digit $2$ raised to the $4$th power.

Execute — Answer: D

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

List every positive divisor of $42$ in order.
Since $42 = 2 \times 3 \times 7$, every divisor is some subset-product of $\{2, 3, 7\}$, giving $2 \times 2 \times 2 = 8$ divisors total.

$$\text{divisors of } 42: 1, 2, 3, 6, 7, 14, 21, 42$$

💡 Listing every divisor in order is exactly the Grade 4 "find all factor pairs" move — every divisor has to show up here.

#5 Look for a Pattern 4.OA.B.4 Step 2

Pair the divisors.
Match each small divisor with its large partner whose product is $42$.
The $8$ divisors break into $4$ partner pairs.

$$1 \cdot 42 = 42, \quad 2 \cdot 21 = 42, \quad 3 \cdot 14 = 42, \quad 6 \cdot 7 = 42$$

💡 Divisors come in pairs whose product is the original number — Tool #5 spots this symmetry and Grade 4 "factor pair" thinking confirms it.

#7 Identify Subproblems 6.EE.A.1 Step 3

Multiply all four pair-products together.
Since each of the $4$ pairs contributes a factor of $42$, the full product $N$ is $42$ raised to the $4$th power.

$$N = 1 \cdot 2 \cdot 3 \cdot 6 \cdot 7 \cdot 14 \cdot 21 \cdot 42 = (1 \cdot 42)(2 \cdot 21)(3 \cdot 14)(6 \cdot 7) = 42^4$$

💡 Four pairs each equaling $42$ multiply to $42^4$ — Grade 6 "evaluate expressions with whole-number exponents".

#7 Identify Subproblems 8.EE.A.1 Step 4

Now find only the units digit of $42^4$.
The ones digit of a power depends only on the ones digit of the base, so the ones digit of $42^4$ equals the ones digit of $2^4$.

$$42^4 \equiv 2^4 \equiv 16 \equiv 6 \pmod{10} \;\Rightarrow\; \textbf{(D)}$$

💡 Tens-and-higher digits cannot affect the ones digit of a product — Grade 8 integer-exponent properties make this routine.

[1] #2 4.OA.B.4 List every positive divisor of $42$ in order. Since $42 = 2 \times 3 \times 7$,

[2] #5 4.OA.B.4 Pair the divisors. Match each small divisor with its large partner whose product

[3] #7 6.EE.A.1 Multiply all four pair-products together. Since each of the $4$ pairs contribute

[4] #7 8.EE.A.1 Now find only the units digit of $42^4$. The ones digit of a power depends only

Review

Reasonableness: Cross-check by computing the units digit the long way. Multiply the ones digits in order: $1 \cdot 2 \cdot 3 \cdot 6 \cdot 7 \cdot 4 \cdot 1 \cdot 2$. Step by step (keeping only ones digits): $1 \to 2 \to 6 \to 6 \to 2 \to 8 \to 8 \to 6$. Ones digit $= 6$, matching answer (D). The pairing trick gives the same answer with far less arithmetic.

Alternative: Tool #2 (Systematic List) alone, skipping the pairing trick: multiply the ones digits left to right as just shown, tracking only the current ones digit. The pairing trick is faster, but the brute list-and-multiply path also lands on $6$, confirming answer (D).

CCSS standards used (min grade 8)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing all $8$ divisors of $42$ and pairing each small divisor $d$ with its partner $42/d$ so every pair multiplies to $42$.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Combining the four pair-products $42 \cdot 42 \cdot 42 \cdot 42$ into the single expression $42^4$.)
8.EE.A.1 Know and apply the properties of integer exponents (Using $42^4 \equiv 2^4 = 16 \pmod{10}$ — the ones digit of a power depends only on the ones digit of the base — to extract the units digit $6$.)

⭐ This AMC 10 problem only needs Grade 8 exponent rules — pair the divisors of $42$ so each pair multiplies to $42$, and the ones digit of $42^4$ is the ones digit of $2^4 = 16$!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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