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M03Q20

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M03Q20 [#permalink]  28 Apr 2009, 14:53
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Which of the following integers is a divisor of $$(15!+13)$$?

(A) 15
(B) 14
(C) 13
(D) 7
(E) 2

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Factorials can be rewritten as follows:
n! = n*(n-1)*(n-2)*(n-3)*...*(n-k)!

Where k can be any number from 1 to n. We can rewrite (15!+13) as follows:
15!+13 = (15*14*13*12!)+13 = 13*(15*14*12!+1)

Therefore, (15!+13) in its final form has a divisor of 13 in it but not 15, 14, 7, or 2.

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Re: M03Q20 [#permalink]  29 Apr 2009, 20:18
13 ( 15*14*12*....1 + 1) can be compared to something like

x(y+1) is divisible by x and not y .So x is a factor of x(y+1) and not y ..
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Re: M03Q20 [#permalink]  01 Feb 2010, 19:14
Alternate Explanation :
15 ! will be divided by even ( because it contains atleast one even)

15! + 13 becomes odd, so strike out (B) and (E)

Also 15 ! will be divided by 15 , so 15 ! + 13 will be 2 short of 15
similarly 7 can be eliminated.
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Re: M03Q20 [#permalink]  18 May 2010, 06:51
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Which of the following integers is a divisor of (15!+13)?

(A) 15
(B) 14
(C) 13
(D) 7
(E) 2

C

explanation

(15!+13) = ((12!*13*14*15)+13) = 13 ((12!*14*15)+1)
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Re: M03Q20 [#permalink]  19 May 2010, 05:48
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The solution is pretty straightforward...!!!

The question asks if (15! + 13) divisible by any of the four choices. I can rewrite this as (a+b)/c = a/c + b/c. Here a = 15!, b=13 and c = one of the four answer choices.

We know 15! will contains all integers from 15 through 1----> so 15! is divisible by all choices.

Now 13 is divisible by only 13 and 1 because it is a prime number.

Therefore, 13 divides both 15! and obviously 13------> HENCE the answer.

Took me about 10 seconds to figure this out.
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Re: M03Q20 [#permalink]  19 May 2010, 09:56
The solution is pretty straightforward...!!!

The question asks if (15! + 13) divisible by any of the four choices. I can rewrite this as (a+b)/c = a/c + b/c. Here a = 15!, b=13 and c = one of the four answer choices.

We know 15! will contains all integers from 15 through 1----> so 15! is divisible by all choices.

Now 13 is divisible by only 13 and 1 because it is a prime number.

Therefore, 13 divides both 15! and obviously 13------> HENCE the answer.

Took me about 10 seconds to figure this out.

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Re: M03Q20 [#permalink]  20 May 2011, 20:28
DLiu1214 wrote:
Hi,

Sorry to sound elementary but I am not familiar with this simplification:

15! + 13 = 13(15*14*12*... + 1)

Can someone explain the general form?

Thank you.

Not really sure what you mean by "general form". The simplification is just the distributive property over additive numbers

$$(x*c + y*c) = (x + y) * c$$

So:

$$15! + 13 = (15 * 14 * 13 * ... * 2 * 1) + 13 = (15 * 14 * 12 * 11 * ... * 2 * 1) * 13 + 1 * 13 = ((15 * 14 * 12 * 11 * ... * 2 * 1) + 1) * 13$$
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Re: M03Q20 [#permalink]  22 May 2011, 16:48
i go with 13.

Reason: 15!/x + 13/x, what is x?
Since 13 is prime it can be divided by 13 and 1. Since 1 not an option and 15! has 13, x=13.
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Re: M03Q20 [#permalink]  30 May 2011, 20:24
While you can use math to explain it simple logic will also suffice.

Talking 15! automatically makes the end result divisible by all the components (so 15! is divisible by 13). So merely adding 13 to this end results guarantees to make it a multiple of 13.

I found it to be really easy (sub 15 sec question) when looked at in that light.
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Re: M03Q20 [#permalink]  03 Mar 2012, 19:09
analyst218 wrote:
15! has three zero's in it. therefore adding 13 to the number would be divisible only by 13.

Isn't this supposed to be the easiest solution? Experts let me know if this method is wrong.

Will have 3 zeroes because we have 3 combinations that can give us a zero 5*2, 10, 15*12
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Re: M03Q20 [#permalink]  23 Mar 2012, 11:43
Why the zeros could be relevant in this case?

For instance,
$$12*13 - 13 = 13*(12 - 1)$$ is divisible by 13 (no zeros!).
$$12*13*10 - 13 = 13*(12*10 - 1)$$ is divisible by 13.
$$12*13*100 - 13 = 13*(12*100 - 1)$$ is divisible by 13.

$$(15! - 13)$$ is divisible by 13 because, although obvious, it is a multiple of 13. You can know that after performing the factorization $$13*(\frac{15!}{13} - 1)$$

Is $$254000 + 13$$ divisible by 13?
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Re: M03Q20 [#permalink]  25 May 2012, 02:45
This question is brilliant as it tests the two concepts of remainder theorm -

1. (Multiplication Rule): This rule is partially applicable here, i would still write it for people who dont know it yet. In an expression (a*b*c)/d, if any of the individual members of the numerator is divisible by d, then the whole numerator is divisible by d. In other words, the remainder of the expression in such a case is zero.

2. (Addition Rule): This one is applicable here. Again in an expression (a+b+c)/d, the remainder will be zero if a, b and c, each, when individually divided by d, gives a remainder zero OR when these individual remainder add up to d itself. Lets take an example to understand better. We all know that 27 is divisible by 9. If i were to write 27 as 9+9+9, then each of the 9's leaves a remainder zero. However if i were to write 27 as 21+2+4, then 21, 2 and 4, when divided by 9, will leave a remainder of 3, 2 and 4 respectively. These remainders add up to 9.

Now lets look at the question: 13! + 13 is always going to be divisible by 13. Why? Apply the fundamentals given above. 13! is nothing but 13*12*11*10 ..... *3*2*1. This is divisible by 13 (Use Multiplication rule on 13! part).

13, the second half, is only divisible by 13 & 1 (It's a prime number). All answer choices EXCEPT 13, leave some remainder. ONLY 13 divides both parts of the expression, leaving an overall remainder zero.

I hope my explanation helped.

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Re: M03Q20 [#permalink]  25 May 2012, 06:45
13 is a multiple of 13.
15! is a multiple of the numbers from 1 to 15 .Hence a multiple of 13.

The sum of multiples of a number is also a multiple of the number.
Since 13 is there in the option, no need to check for any other options.
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Re: M03Q20 [#permalink]  30 May 2012, 16:31
15! will be divided by every integer till 15 , the question is what integer divides 13, which is 13 itself. Hence C.
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Re: M03Q20 [#permalink]  24 May 2013, 02:48
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hey
the 15! = (15*14*13*12*11*10!)
so it is clear that 13 is a divisor of 15!, and adding 13 to it makes the quotient increase by 1
thus 13 is a divisor of (15!+13) .

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Re: M03Q20 [#permalink]  24 May 2013, 02:57
Expert's post
If $$n=15!+13$$, which of the following is a divisor of $$n$$?

A. 15
B. 14
C. 13
D. 7
E. 2

$$15!=1*2*3*...12*13*14*15$$, so $$n=1*2*3*...12*13*14*15+13$$ --> factor out 13: $$n=13*(1*2*3*...12*14*15+1)$$, hence 13 is definitely a divisor of $$n$$.

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Re: M03Q20 [#permalink]  24 May 2013, 04:12
!15 and 13 , both have 13 as their factors, so 13 is the answer
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Re: M03Q20 [#permalink]  24 May 2013, 05:27
shakeandbake wrote:
Which of the following integers is a divisor of $$(15!+13)$$?

(A) 15
(B) 14
(C) 13
(D) 7
(E) 2

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Factorials can be rewritten as follows:
n! = n*(n-1)*(n-2)*(n-3)*...*(n-k)!

Where k can be any number from 1 to n. We can rewrite (15!+13) as follows:
15!+13 = (15*14*13*12!)+13 = 13*(15*14*12!+1)

Therefore, (15!+13) in its final form has a divisor of 13 in it but not 15, 14, 7, or 2.

(15!= 15x14x13x12x11x.......x2x1) is a multiple of 13 . When you add 13 to a multiple of 13, it is still a multiple of 13 and divisible by 13. So answer is C.
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Re: M03Q20 [#permalink]  27 May 2013, 20:28
Just curious, would something like this be a 600-level question?
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Re: M03Q20 [#permalink]  27 May 2013, 23:45
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Expert's post
Dixon wrote:
Just curious, would something like this be a 600-level question?

Yes, I'd say it's a sub-600 question.
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Re: M03Q20   [#permalink] 27 May 2013, 23:45

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