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

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(→‎🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️))
 
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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
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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>
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==Solution 1==
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The factors of <math>42</math> are: <math>1, 2, 3, 6, 7, 14, 21, 42</math>. Multiply 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> which 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} = 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>\boxed{\textbf{(D) } 6}</math>.
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==🎥✨ Video Solution by 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 1 by Pi Academy (Fast and Easy ⚡🚀)==
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https://youtu.be/QLziG_2e7CY?feature=shared
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~ Pi Academy
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
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==See also==
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{{AMC10 box|year=2024|ab=B|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 06:33, 19 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 $\boxed{\textbf{(D) } 6}$

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 $42 = 2^1 \cdot 3^1 \cdot 7^1$ 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 $\boxed{\textbf{(D) } 6}$.

🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)

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

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/QLziG_2e7CY?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

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

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

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