SOLUTION: We choose a positive divisor of 20^{20} at random (with all divisors equally likely to be chosen). What is the probability that we chose a multiple of 5?

Question 1210203: We choose a positive divisor of 20^{20} at random (with all divisors equally likely to be chosen). What is the probability that we chose a multiple of 5?
Found 3 solutions by CPhill, greenestamps, math_tutor2020:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Let's break down this problem step by step.
**1. Prime Factorization of 20^20**
* 20 = 2² * 5
* 20²⁰ = (2² * 5)²⁰ = 2⁴⁰ * 5²⁰
**2. Number of Divisors**
The number of divisors of 20²⁰ is found by adding 1 to each exponent in the prime factorization and multiplying them:
* Number of divisors = (40 + 1) * (20 + 1) = 41 * 21 = 861
**3. Divisors that are Multiples of 5**
For a divisor to be a multiple of 5, it must have at least one factor of 5.
* The form of a divisor of 20²⁰ is 2^a * 5^b, where 0 ≤ a ≤ 40 and 0 ≤ b ≤ 20.
* For a multiple of 5, we need b ≥ 1, so 1 ≤ b ≤ 20.
* The number of choices for 'a' is still 41 (0 to 40).
* The number of choices for 'b' is 20 (1 to 20).
* Number of multiples of 5 = 41 * 20 = 820
**4. Calculate the Probability**
* Probability = (Number of multiples of 5) / (Total number of divisors)
* Probability = 820 / 861
**5. Simplify the Fraction**
We can simplify the fraction by dividing both numerator and denominator by their greatest common divisor.
* 820 = 2² * 5 * 41
* 861 = 3 * 7 * 41
The greatest common divisor is 41.
* Probability = (820 / 41) / (861 / 41)
* Probability = 20 / 21
**Therefore, the probability that we chose a multiple of 5 is 20/21.**

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

The prime factorization of 20 is , so the prime factorization of 20^20 is .

If a positive divisor is a multiple of 5, then it contains at least one factor of 5.

The divisors of 20^20 that are NOT multiples of 5 are those that contain only factors of 2.

To count the number of positive divisors of a number, add 1 to each exponent in the prime factorization of the number and multiply the resulting numbers.

The number of divisors of 20^20 is

The number of divisors of 20^20 that contain only factors of 2 is

The number of divisors of 20^20 that DO contain at least one factor of 5 is 861-41 = 820.

The probability that a random divisor of 20^20 is a multiple of 5 is

ANSWER: 20/21

Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Refer to the scratch work that tutor greenestamps has written out.

Note that he mentioned
(40+1)(20+1) = 41*21 = 861
However, it is best to leave it as 41*21
If you're curious how this formula works, then refer to my response on this page
Additionally you can refer to formula (3) of this page

A = number of divisors of 20^20
B = number of divisors of 20^20 that do not involve 5 (i.e. involve 2's only)
C = number of divisors of 20^20 that have at least one factor of 5

A = 41*21
B = 41

C = A-B
C = 41*21 - 41
C = 41*(21 - 1)
C = 41*20

D = probability a divisor of 20^20 has at least one factor of 5
D = C/A
D = (41*20)/(41*21)
D = 20/21 is the final answer in fraction form.

20/21 = 0.95238 = 95.238% approximately
Round this approximate value however needed.

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