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

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==Problem==
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== Problem ==
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Let <math>N</math> be the product of all the positive integer divisors of <math>42</math>. What is the units digit
 
Let <math>N</math> be the product of all the positive integer divisors of <math>42</math>. What is the units digit
 
of <math>N</math>?
 
of <math>N</math>?
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<math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8</math>
 
<math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8</math>
  
==Solution 1==
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== Solution 1 ==
The factors of <math>42</math> are: <math>1, 2, 3, 6, 7, 14, 21, 42</math>. Multiply unit digits to get D) 6
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The factors of <math>42</math> are <math>1, 2, 3, 6, 7, 14, 21, 42</math>. Multiply the unit digits to get <math>\boxed{\textbf{(D) } 6}</math>
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==Solution 2==
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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> has <math>(1+1)(1+1)(1+1) = 8</math> factors, so the product of the divisors of <math>42</math> is
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<cmath>42^{\frac{8}{2}} \equiv 42^4 \equiv 2^4 \equiv 16 \equiv \boxed{\textbf{(D) } 6} \pmod{10}.</cmath>
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== Video Solution by 1 Scholars Foundation (Easy to Understand)==
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https://youtu.be/T_QESWAKUUk?si=E8c2gKO-ZVPZ2tek&t=201
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== Video Solution 2 by Pi Academy (Fast and Easy) ==
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https://youtu.be/QLziG_2e7CY
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==Video Solution 3 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
  
==See also==
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== 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}}

Latest revision as of 12:46, 21 January 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

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


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

See Also

2024 AMC 10B (ProblemsAnswer KeyResources)
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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