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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 10^18?

 Feb 16, 2022
 #1
avatar+516 
+6

20^20 prime factorized is 2^40 x 5^20. Thus, using the prime factorization trick to find the # of factors, 20^20 has 861 factors. 

 

10^18 prime factorszed is 2^18 x 5^18. Thus, using the prime factorization trick to find the # of factors, 10^18 has 361 factors. 

 

Since all factors greater than 10^18 in 20^20 are a multiple of 10^18, thus there are 861 - 361 + 1 = 501 factors that are a multiple of 10^18 when choosing one from 20^20. 

(The reason I had a "+ 1" is because one factor of 10^18 is 10^18, which is also a multiple of 10^18. Thus there are 360 factors that do not work in 20^20.)

That means, the probability of choosing a divisor from 20^20 and having it be a multiple of 10^18 is 501/861.

 

Simplifying, we get the probability as 167/287.

smiley

 Feb 16, 2022
 #2
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+1

 

You started on the right track and then went astray!

 

[20^20] / [10^18] ==104,857,600==2^22  x  5^2 ==which has [22 + 1] x [2 + 1]==23  x  3==69 divisors.

 

20^20 ==1.048576e+26 = 2^40 * 5^20 ==41 * 21 ==861 divisors.

 

The probability is: 69 / 861 == 23 / 287

 Feb 16, 2022

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