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Re: If n = 20! + 17, then n is divisible by which of the following? [#permalink]
BrentGMATPrepNow wrote:
Kimberly77 wrote:

Hi Brent BrentGMATPrepNow, thanks for the reply.
To clarify, 17 is being factored out like below and rest of number are some number + 1. So how is this part some number + 1 from multiplication with 17 become quotient when divide by 17? Have I missed something here?

20! +17 = 17(2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20+1)


Let's look at some examples:
We know that (3)(4) = 12. Notice that if we divide 12 by 3, the quotient is 4 and the remainder is 0. This means we can say 12 is divisible by 3 and 4.
Similarly, we know that (5)(7) = 35. Notice that if we divide 35 by 7, the quotient is 5 and the remainder is 0. This means we can say 35 is divisible by 5 and 5.

Similarly, 20! + 17 = (17)(some number + 1). So, if we divide 20! + 17 by 17, the quotient is (some number + 1) and the remainder is 0. This means we can say 20! + 17 is divisible by 17


Great explanation BrentGMATPrepNow and you rock :thumbsup: :thumbsup: . Thanks Brent and crystal clear now :please:
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Re: If n = 20! + 17, then n is divisible by which of the following? [#permalink]
Bunuel wrote:
Walkabout wrote:
GENERALLY:
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope it helps.


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?
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If n = 20! + 17, then n is divisible by which of the following? [#permalink]
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.
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If n = 20! + 17, then n is divisible by which of the following? [#permalink]
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