If n = 20! + 17, then n is divisible by which of the following?

Great explanation BrentGMATPrepNow and you rock . Thanks Brent and crystal clear now

Bunuel

What if k is an integer and k < 0 and k is a factor of integers a and b? do none of the above rules apply?

Imagine you have a huge number made by multiplying all the numbers from 1 to 20 together. That's what we call 20 factorial, or \(20!\) for short. Now, if we add 17 to that huge number, we get a new number, let's call it \(n\).

Now, we're trying to figure out if certain numbers can evenly divide \(n\) without leaving any leftovers. These numbers are 15, 17, and 19.

1. **For 15**: To see if a number can be divided by 15 without leftovers, it needs to be evenly divisible by both 3 and 5. Our huge number from 1 to 20 includes both 3 and 5, so it can be divided by 15. But, when we add 17 to it, it doesn't work out to be evenly divisible by 15 anymore. So, \(n\) doesn't work with 15.

2. **For 17**: Since we're adding 17 to our huge number that already includes a 17, it kind of makes it a special case. It turns out that adding 17 doesn't stop \(n\) from being divisible by 17. So, \(n\) works with 17!

3. **For 19**: Just like with 15, our huge number includes 19, but adding 17 doesn't help it to be divided by 19 without leftovers. So, \(n\) doesn't work with 19.

So, after checking, only the number 17 can divide \(n\) evenly without any leftovers.

Multiple + Multiple = Multiple
Multiple + Non-Multiple = Non-Multiple
Non-Multiple + Non-Multiple = ?

20! is a multiple of all numbers 1-20. 17 is restricted to only 17 since it is a prime. Thus only 17 can be a factor of 20! + 17