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

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

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.

@@ Line 1: / Line 1: @@
-If Bob billy and bilal all eat 2, 3, and 912 watermelons respectively, whats the best way they can eat all of them such that only half of the watermelons have integer divisors less than 2?
+==Problem==
+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>?
+<math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8</math>
+==Solution 1==
+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>
+==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{\textbf{(D) } 6}</math>.
+==🎥✨ 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==
+{{AMC10 box|year=2024|ab=B|num-b=7|num-a=9}}
+{{MAA Notice}}

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