Determine the largest value of a positive integer n, such that:
25 divides n!+1
The first thought is that 24+1 = 25 and 24 is 4!, so n=4 is a possibility.
The second thought is that 25 divides numbers that end in two zeros and does not divide numbers that end in 01. Since 10! ends with 2 zeros, n < 10.
Third thought:
n! must end in {24,49,74,99}
But for the last 2 digits of a factorial:
- 49 is out because it has 2 prime factors of 7,
- 74 is out because it has a prime factor of 37 (>10), and
- 99 is out because it has a prime factor of 11 (>10).
So n! must end in 24.
But (fourth thought) 5! (or any larger factorial) must end in either 0 or 5, not 24.
So n=4 is the only solution.
(really just needed the first and fourth thoughts)
|
|
Posted by Larry
on 2024-08-27 12:15:42 |
solution
