Which of the following integers is a divisor of (15!+13)?

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


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.

I'm having a difficult time understanding this explanation. Can anyone please help me understand. Thank you in advance.

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)

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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C

15!=1*2*3*...*13*14*15
15! + 13 = 13 (1*2*3*...*1*14*15 + 1), divisible by 13

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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My Debrief

15! is already divisible by 1 to 15 .

so 15!+13 will be divisible by only 13

option C
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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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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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15! will be divided by every integer till 15 , the question is what integer divides 13, which is 13 itself. Hence C.

!15 and 13 , both have 13 as their factors, so 13 is the answer