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Difference between revisions of "2024 AMC 10B Problems/Problem 8"

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<cmath>42^\frac{8}{2} = 42^4.</cmath>
 
<cmath>42^\frac{8}{2} = 42^4.</cmath>
  
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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>\textbf{(D) } 6</math>.
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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{\textbf{(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 03:26, 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 $\boxed{\textbf{(D) } 6}$.

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

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