2024 AMC 10B Problems/Problem 8

Problem

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

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

Solution 1

The factors of $42$ are: $1, 2, 3, 6, 7, 14, 21, 42$ . Multiply unit digits to get D) 6

Solution 2

The product of the factors of a number $n$ is $n^\frac{\tau(n)}{2}$ , where $\tau(n)$ is the number of positive divisors of $n$ . We see that $42 = 2^1 \cdot 3^1 \cdot 7^1$ which has $(1+1)(1+1)(1+1) = 8$ factors, so the product of the divisors of $42$ is

$42^\frac{8}{2} = 42^4.$

But we only need the last digit of this, which is the same as the last digit of $2^4$ . The answer is $\boxed{\textbf{(D) } 6}$ .

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2024 AMC 10B Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

See also