https://prepswift.gregmat.com/quizzes/quiz/prepswift-of-numbers-in-factorials-iii (Question 6)
write your question here…I don’t quite understand the solution to this, can someone help simply the solution?
What do you understand?
I dont understand what the question is asking. I understand everything until Greg states "Can you see where we’re going from here? If n+1n+1 is divisible by 7373, three extra 77s come out (from 77, 7272 and 7373), which means that the largest integer pp also increases by 33.
As a result, n+1=73n+1=73 or n=343−1=342n=343−1=342. This is the smallest value of nn since any non-zero multiple of 343343 will work for n+1n+1. So another answer would be 2×343−1=6852×343−1=685, albeit the second-smallest one." as the solution.
Take an easier example then we can segue into the actual question.
What’s the highest p possible for \frac{\text{7!}}{\text{7}{^p}} to be an integer?
>> Ans is 1
What if I do \frac{\text{(7+1)!}}{\text{7}{^p}}
>> Ans is still 1
Side Note: I chose 7! as a simple test case given I’d be dividing by 7^p, you could have chosen any other number.
Applying that same logic to the question,
What is the highest possible p for \frac{\text{n!}}{\text{7}{^p}} to be an integer?
>> Ans is x according to question.
What if I do \frac{\text{(n+1)!}}{\text{7}{^p}}
>> Ans is (x+3) according to question.
So, we need a number, n that:
- Its factorial, n! has x 7s
- And (n+1)! has (x+3) 7s
342 is that number, n because 343 = 7³ introduces three 7s when 342 increases by 1 to become 343.