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I have just solved two similar assignments and I am confused about the answer guides. Both assignments require the same solution method, however, one answer guide says the solution must be multiplied by the number of digits in the numerator, and the other doesn't. So for the 20 factorial assignment, we must multiply the answer of 4 by two and for the 500 factorial, we do not need to multiply the 124. What is the reasoning?
Problem 1:
Solution:
We are given that 20!×20!/20n is an integer, and we need to determine the largest possible integer value of n. Breaking 20n20^n20n into prime factors, we have:
We see that since there are fewer 5’s than 2’s in 20!20!20!, the greatest possible integer value of n is dependent on the number of 5’s in 20!20!20!. To determine the number of 5’s in 20!20!20!, we use the factorial division shortcut:
20/5 = 4
We see that there are 4 fives in the first 20!20!20! and another 4 fives in the second 20!20!20!. In other words, there are 8 fives in 20!×20!20! \times 20!20!×20!. Thus, the largest integer value of n is 8.
Problem 2:
Solution:
We are given that 500!×500!×500!/500n is an integer, and we need to determine the largest possible integer value of n.
In other words, the question is asking us to perform the factorial division shortcut:
We see that there are 100+20+4=124 fives in 500!. Thus, there are another 124 fives in the next 500!500!500! and another 124 fives in the third 500!500!500!. In other words, there are 124×3=372.
We must remember that we need to determine the maximum integer value of n such it is an integer. Thus:
3n≤372 --> n≤124
Therefore, the largest integer value of n is 124.
In problem 1, why do I not need to do 2n < 8 as in the solution guide for problem 2? THANK YOU to anyone who can help!!
Problem 1:
Solution:
We are given that 20!×20!/20n is an integer, and we need to determine the largest possible integer value of n. Breaking 20n20^n20n into prime factors, we have:
We see that since there are fewer 5’s than 2’s in 20!20!20!, the greatest possible integer value of n is dependent on the number of 5’s in 20!20!20!. To determine the number of 5’s in 20!20!20!, we use the factorial division shortcut:
20/5 = 4
We see that there are 4 fives in the first 20!20!20! and another 4 fives in the second 20!20!20!. In other words, there are 8 fives in 20!×20!20! \times 20!20!×20!. Thus, the largest integer value of n is 8.
Problem 2:
Solution:
We are given that 500!×500!×500!/500n is an integer, and we need to determine the largest possible integer value of n.
In other words, the question is asking us to perform the factorial division shortcut:
We see that there are 100+20+4=124 fives in 500!. Thus, there are another 124 fives in the next 500!500!500! and another 124 fives in the third 500!500!500!. In other words, there are 124×3=372.
We must remember that we need to determine the maximum integer value of n such it is an integer. Thus:
3n≤372 --> n≤124
Therefore, the largest integer value of n is 124.
In problem 1, why do I not need to do 2n < 8 as in the solution guide for problem 2? THANK YOU to anyone who can help!!
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