2021 Fall AMC 10B Problems/Problem 6
Contents
Problem
The least positive integer with exactly distinct positive divisors can be written in the form , where and are integers and is not a divisor of . What is
Solution 1
Let this positive integer be written as . The number of factors of this number is therefore , and this must equal 2021. The prime factorization of 2021 is , so and . To minimize this integer, we set and . Then this integer is . Now and so
~KingRavi
Solution 2
Recall that can be written as . Since we want the integer to have divisors, we must have it in the form , where and are prime numbers. Therefore, we want to be and to be . To make up the remaining , we multiply by , which is which is . Therefore, we have
~Arcticturn
Solution 3
If a number has prime factorization , then the number of distinct positive divisors of this number is .
We have . Hence, if a number has 2021 distinct positive divisors, then takes one of the following forms: , .
Therefore, the smallest is .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=530
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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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