2024 AMC 10B Problems/Problem 8: Difference between revisions

Revision as of 20:01, 23 April 2025

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 the unit digits to get $\boxed{\textbf{(D) } 6}$

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 $42 = 2^1 \cdot 3^1 \cdot 7^1$ has $(1+1)(1+1)(1+1) = 8$ factors, so the product of the divisors of $42$ is

$42^{\frac{8}{2}} \equiv 42^4 \equiv 2^4 \equiv 16 \equiv \boxed{\textbf{(D) } 6} \pmod{10}.$

Solution 3: Product of Factors, Easy to Understand

Product of factors : 42 ^ amount of factors/2

3*2*7 so that leads to 8 factors.

Units digit of 42^4 is what we need to find.

2^4 ends in 6 so \boxed{\textbf{(D) } 6}

Video Solution by 1 Scholars Foundation (Easy to Understand)

https://youtu.be/T_QESWAKUUk?si=E8c2gKO-ZVPZ2tek&t=201

Video Solution 2 by Pi Academy (Fast and Easy)

https://youtu.be/QLziG_2e7CY

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

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
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Page

Toolbox

Search