Difference between revisions of "2024 AMC 10B Problems/Problem 8"
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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 D) 6 | The factors of <math>42</math> are: <math>1, 2, 3, 6, 7, 14, 21, 42</math>. Multiply unit digits to get D) 6 | ||
+ | ==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{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 02:48, 14 November 2024
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 unit digits to get D) 6
Solution 2
The product of the factors of a number is , where is the number of positive divisors of . We see that which has factors, so the product of the divisors of is
But we only need the last digit of this, which is the same as the last digit of . The answer is .
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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