Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 10B Problems/Problem 8"
Line 8: Line 8:
 
The factors of <math>42</math> are: <math>1, 2, 3, 6, 7, 14, 21, 42</math>. Multiply unit digits to get D) 6
 
The factors of <math>42</math> are: <math>1, 2, 3, 6, 7, 14, 21, 42</math>. Multiply unit digits to get D) 6
  
 +
==Solution 2==
 +
The product of the factors of a number <math>n</math> is <math>n^\frac{\tau(n)}{2}</math>, where <math>\tau(n)</math> is the number of positive divisors of <math>n</math>. We see that <math>42 = 2^1 \cdot 3^1 \cdot 7^1</math> which has <math>(1+1)(1+1)(1+1) = 8</math> factors, so the product of the divisors of <math>42</math> is
 +
 +
<cmath>42^\frac{8}{2} = 42^4.</cmath>
 +
 +
But we only need the last digit of this, which is the same as the last digit of <math>2^4</math>. The answer is <math>\boxed{D (6)}</math>.
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=B|num-b=7|num-a=9}}
 
{{AMC10 box|year=2024|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:48, 14 November 2024

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 $\boxed{D (6)}$.

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10B_Problems/Problem_8&oldid=233709"