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Find the largest integer n for which 12^n evenly divides 20! + 12!.

 Jul 22, 2022
 #1
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20! + 12! = 12!(x + 1). Since we know that x is an even number, (x + 1) is an odd number, and therefore cannot "produce" any more factors of 12.

Therefore, the largest integer n for which 12^n divides it would be how much 12^n divides 12!.

 

12 = 2^2 * 3

 

We have to see how many factors of 2 are there in 12!, divide that by 2 and set that equal to y.

In addition, we can see how many factors of 3 there are in 12! and set that equal to z. 

 

1: 2, 4, 6, 8, 10, 12

2: 4, 8, 12

3: 8

 

Adding up gives us: 10

Therefore, there are 5 2^2 terms.

 

1: 3, 6, 9, 12

2: 9

There are also 5 factors of 3 in the expression.

Therefore, 5 is the largest integer for n that divides 20! + 12!

 Jul 22, 2022
 #2
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Since 12 ==2^2  x  3, we can see that the largest distinct factor of 12==3, and since:

 

20!  +  12!= 2^10 * 3^5 * 5^2 * 7 * 11 * 23 * 31 * 653 * 10909

 

Therefore, the largest exponent of 3 above==5 - the largest integer n, and:

 

[20!  +  12!]  mod  12^5 ==0

 Jul 22, 2022
 #3
avatar+197 
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It's absolutely ridiculous to not only solve for 20! and 12! but to also take the prime factorization!

 Jul 22, 2022

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