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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
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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Walkabout
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.
Notice that we can factor out 17 from from 20! + 17 (both 20! and 17 are divisible by 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.
Check this also:
20! IS divisible by 15 but 17 is NOT. Thus 20! + 17 = (a multiple of 15) + (NOT a multiple of 15) = (NOT a multiple of 15). So, 20! + 17 is NOT divisible by 17.
20! IS divisible by 19 but 17 is NOT. Thus 20! + 17 = (a multiple of 19) + (NOT a multiple of 19) = (NOT a multiple of 19). So, 20! + 19 is NOT divisible by 19.
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.
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Here its how it works we are given 20*19*18....*1 + 17
option 1 15 now if we try to divide by 15 the first part is 20!/15 ( this part is divisible) + 17/15 ( this isn't)
option 2 17 we can take a 17 common and then its divisible
option 3 the first part is 20!/19 ( this part is divisible) + 17/19 ( not divisble)
hence option C.
option 1 15 now if we try to divide by 15 the first part is 20!/15 ( this part is divisible) + 17/15 ( this isn't)
option 2 17 we can take a 17 common and then its divisible
option 3 the first part is 20!/19 ( this part is divisible) + 17/19 ( not divisble)
hence option C.
Posts: 6,727
Walkabout
Answer choice I: is 20! + 17 divisible by 15?
20! + 17 = (20)(19)(18)(17)(16)(15)(other stuff) + 15 + 2
= (15)(some number + 1) + 2
(15)(some number + 1) is a multiple of 15
So, (15)(some number + 1) + 2 is 2 greater than a multiple of 15
So, if we divide (15)(some number + 1) + 2 by 15, the remainder will be 2
So, 20! + 17 is NOT divisible by 15
ELIMINATE B and D
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
So, 20! + 17 IS divisible by 17
ELIMINATE A
Answer choice III: is 20! + 17 divisible by 19?
20! + 17 = (20)(19)(other stuff) + 17
= (19)(some number) + 17
If we divide (19)(some number) + 17 by 19, the remainder will be 17
So, 20! + 17 is NOT divisible by 19
ELIMINATE E
Answer: C
Cheers,
Brent
General Discussion
Walkabout
Another point to note here is that if a number n is divisible by m (where m is greater than 1), (n+1) will not be divisible by m.
So 20! + 17 = 17*(1*2*3*...15*16*18*19*20 + 1)
Assume 1*2*3*...15*16*18*19*20 = N
So 20! + 17 = 17(N + 1)
If this number has to be divisible by either 15 or 19, (N+ 1) must be divisible by 15 or 19.
Since N is divisible by both 15 and 19, (N + 1) can be divisible by neither.
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But in this case !20+17 is not completely divisible by 17.So 17 is not the factor of sum of !20+17.Please clarify?
anu1706
Actually it is:
\(20! + 17=17(1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20+1)\)
\(\frac{20! + 17}{17}=1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20+1\)
Hope it's clear.
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Is this type of problem similar to remainder problems? My thinking was that n = 20(n) + 17 with zero remainder, meaning it had to be divisible by 17. If it were divisible by 17, then it was not divisible by 15 or 19. Does that make sense though?
TAL010
I am not sure I understand what you did there.
How do you get n = 20a + 17 has to be divisible by 17? (I am assuming you meant the second variable to be different)
The reason 20! + 17 must be divisible by 17 is that 20! = 1*2*3*4*...17*18*19*20
So 20! is a multiple of 17 and can be written as 17a
n = 17a + 17 is divisible by 17.
Also, a number can be divisible by 17 as well as 15 as well as 19. e.g. n =15*17*19 is divisible by all three.
But here n = 20! + 17 in which 20! is divisible by 15 as well as 19 but 17 is neither divisible by 15 nor by 19. So when you divide n by 15, you will get a remainder of 2. When you divide n by 19, you will get a remainder of 17.
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I was thinking that n = 20! + 17 -- knowing this you know that 20! is a multiple of all of the numbers 15, 17 and 19, because 20! = 20x19x18x17x16x15...etc.etc.
However by adding 17, the number is no longer a multiple of 15 or 19 because 17 is not divisible by 19 or 15.
i
However by adding 17, the number is no longer a multiple of 15 or 19 because 17 is not divisible by 19 or 15.
i
VeritasPrepKarishma
TAL010
Yes, this is fine. Note that previously, you had written
n = 20a + 17 is divisible by 17. I am assuming the 'a' was a typo.
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ScottTargetTestPrep
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Expert reply
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Walkabout
We are given that n = 20! + 17 and need to know whether n is divisible by 15, 17, and/or 19. To determine this, we rewrite the given expression for n using each answer choice.
Thus, we have:
Does (20! + 17)/15 = integer?
Does (20! + 17)/17 = integer?
Does (20! + 17)/19 = integer?
We now use the distributive property of division over addition to determine which of these expressions is/are equal to an integer.
The distributive property of division over addition tells us that (a + c)/b = a/b + c/b. We apply this rule as follows:
I.
Does (20! + 17)/15 = integer?
Does 20!/15 + 17/15 = integer?
Although 20! is divisible by 15, 17 is NOT, and thus (20! + 17)/15 IS NOT an integer.
We can eliminate answer choices B and D.
II.
Does (20! + 17)/17 = integer?
Does 20!/17 + 17/17 = integer?
Both 20! and 17 are divisible by 17, and thus (20! + 17)/17 IS an integer.
We can eliminate answer choice A.
III.
Does (20! + 17)/19 = integer?
Does 20!/19 + 17/19 = integer?
Although 20! is divisible by 19, 17 is NOT, so (20! + 17)/19 IS NOT an integer.
We can eliminate answer choice E.
Thus, II is the only correct statement.
Answer: C
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Bunuel
Bunuel, Greetings!
What is the point of taking "17" out of brackets? what is the point of factoring here? How do we get "1" ? Also I am still trying to understand if 20! includes both "15" and 19" than why 20!+17 isn't divisible by "15" and 19". So what if there are two "17" s, ....."15" and 19" are still in 20! 
thanks for taking time to explain and have a great day:)
dave13
20! + 17 = 17*(1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20 + 1), so it's divisible by 17.
Next, pleas re-read the highlighted part carefully.
20! IS divisible by 15 but 17 is NOT. Thus 20! + 17 = (a multiple of 15) + (NOT a multiple of 15) = (NOT a multiple of 15);
20! IS divisible by 19 but 17 is NOT. Thus 20! + 17 = (a multiple of 19) + (NOT a multiple of 19) = (NOT a multiple of 19).
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Bunuel
In 20!,we have 15,17, 19 included in the value.
Therefore it is divisible by all of them. But 20!+17 will be divisible by 17 only.
C.
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Hi Bunuel,
Quick question: Is it a hard rule in which n!, regardless of the number n itself, if you add 1 to n!, none of the numbers would be a divisor of that number?
e.g., if n=4! then n + 1 is NOT divisible by 4, 3, 2, or 1. Similarly, for n! = 20! +17, once you factor out 17 and get 20(19)(18)...(2)(1)+1, this number would NOT be divisible by any of the numbers that are within that product? Is this logic correct? Pretty sure I read it somewhere but I can't seem to find the source..
Quick question: Is it a hard rule in which n!, regardless of the number n itself, if you add 1 to n!, none of the numbers would be a divisor of that number?
e.g., if n=4! then n + 1 is NOT divisible by 4, 3, 2, or 1. Similarly, for n! = 20! +17, once you factor out 17 and get 20(19)(18)...(2)(1)+1, this number would NOT be divisible by any of the numbers that are within that product? Is this logic correct? Pretty sure I read it somewhere but I can't seem to find the source..
samgyupsal
Consecutive integers are co-prime - they do not share any common factor but 1 (x and x + 1 (where x is a positive integer) do not share any common factor but 1). For example, 25 and 26 are consecutive integers and the only common factor they share is 1.
20! +17 = 17(2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20+1)
Here 2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20+1 will not bi divisible by any factor of 2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*18*19*20 because of the rule above.
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Walkabout
Answer: Option C
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BrentGMATPrepNow
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
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Kimberly77
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
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