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

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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 . Thanks Brent and crystal clear now
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Re: If n = 20! + 17, then n is divisible by which of the following? [#permalink]
Bunuel 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.
If n = 20! + 17, then n is divisible by which of the following? [#permalink]
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