Difference between revisions of "2024 AMC 10B Problems/Problem 8"
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| − | + | == 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 the 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> 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}} \equiv 42^4 \equiv 2^4 \equiv 16 \equiv \boxed{\textbf{(D) } 6} \pmod{10}.</cmath> | ||
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| + | |||
| + | |||
| + | ==Solution 3: Product of Factors, Easy to Understand== | ||
| + | |||
| + | Product of factors : 42 ^ amount of factors/2 | ||
| + | |||
| + | 3*2*7 so that leads to 8 factors. | ||
| + | |||
| + | Units digit of 42^4 is what we need to find. | ||
| + | |||
| + | 2^4 ends in 6 so we get 6. | ||
| + | |||
| + | ~Aarav22 | ||
| + | |||
| + | == 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 == | ||
| + | {{AMC10 box|year=2024|ab=B|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:03, 23 April 2025
Contents
[hide]Problem
Let 
be the product of all the positive integer divisors of 
. What is the units digit
of ?
Solution 1
The factors of are 
. Multiply the unit digits to get 
Solution 2
The product of the factors of a number 
is 
, where 
is the number of positive divisors of . We see that 
has 
factors, so the product of the divisors of is
![\[42^{\frac{8}{2}} \equiv 42^4 \equiv 2^4 \equiv 16 \equiv \boxed{\textbf{(D) } 6} \pmod{10}.\]](http://latex.artofproblemsolving.com/a/b/7/ab7c093ae411c356bc62c21fb065b0d045c260d0.png)
Solution 3: Product of Factors, Easy to Understand
Product of factors : 42 ^ amount of factors/2
3*2*7 so that leads to 8 factors.
Units digit of 42^4 is what we need to find.
2^4 ends in 6 so we get 6.
~Aarav22
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)
Video Solution 3 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
