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

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Kimberly77

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25 Nov 2022, 14:58

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BrentGMATPrepNow

Kimberly77

Hi Brent BrentGMATPrepNow , in Answer choice II: is 20! + 17 divisible by 17?
20! + 17 = (20)(19)(18)(17)(other stuff) + 17
= (17)(some number + 1)
If we divide (17)(some number + 1) by 17, the remainder will be 0 ? >> not sure I quite understand this part? 17 will cancel out but we still have (some number + 1) left behind right?
So, not sure how is 20! + 17 IS completely divisible by 17 for (some number + 1) part?

Thanks Brent

I think you may be confusing quotient with remainder.

If I divide 51 by 17, the quotient is 3 and the remainder is 0.
Since the remainder is 0, I know that 51 is divisible by 17.

Similarly, if we divide (17)(some number + 1) by 17, the quotient will be some number + 1, and the remainder will be 0.
Since the remainder is 0, we know that 20! + 17 is divisible by 17

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)

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25 Nov 2022, 16:32

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Walkabout

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

I. 15
II. 17
III. 19

(A) None
(B) I only
(C) II only
(D) I and II
(E) II and III

n is not divisible by 15, because 20! is divisible by 15 but 17 is not. Similarly, n is not divisible by 19 because 20! is divisible by 19 but 17 is not. On the other hand, since 20! and 17 are both divisible by 17, n = 20! + 17 is divisible by 17.

Answer: C

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26 Nov 2022, 08:44

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Kimberly77

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

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29 Feb 2024, 08:16

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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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28 Oct 2024, 12:29

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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

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18 Aug 2025, 04:17

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That's why easy questions are important, they cement your concepts.

n = 20! + 17
lets make it.... 17(1x2x3....x16x18x19x20 + 1)
So its certainly divisible by 17...

Now, for 15...remember if a is a factor of n then a won't be a factor of n+1...Consider this number... 1x2x3....x16x18x19x20..Is 15 a factor of this? yes..So if we add 1 to this 15 won't be factor..hence not divisible...So 17 also not divisble by 15 and this n+1 also not divisble by 15..
Same for 19 also....

Hence..C..II only...

Walkabout

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

I. 15
II. 17
III. 19

(A) None
(B) I only
(C) II only
(D) I and II
(E) II and III

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10 Dec 2025, 02:18

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Bunuel

20! is the product of all integers from 1 to 20, inclusive, thus it's divisible by each of the integers 15, 17, and 19.

Next, notice that we can factor out 17 from from 20! + 17, thus 20! + 17 is divisible by 17 but we cannot factor out neither 15 nor 19 from 20! + 17, thus 20! + 17 is not divisible by either of them.

Answer: C.

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.

I'm curious if the solution works with other cases as well. For example, if n = 2! + 10 is n divisible by 3? --> (2!+10)/3 = 12/3 = 4; so 2! + 10 is actually divisible by 3 even though 2! isn't divisible by 3 and 10 isn't divisible by 3 individually. For I., I wonder how we can be sure that even if 20! isn't divisible by 15 and 17 isn't divisible by 15, it's not going to end up being divisible by 15 just like the example I gave earlier. The same question goes for III. also.

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10 Dec 2025, 04:50

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thanadol.r

The point here is that 20! itself is a multiple of 15, so for 20! + x to be divisible by 15, x must also be a multiple of 15. In your example of 2! + 10, 2! is not divisible by 3, so it doesn't apply to this case.

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.

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