Find the number that divides 16!+1?

Question

A. 7
B. 11
C. 17
D. 6
E. 18

Bunuel · Accepted Answer

16!  and  16!+1  are consecutive integers. Consecutive integers are co-prime, which means that they do not share any common factor but 1.

Now, obviously  16!  has 6, 7, 11, and 18 as its factors so  16!+1  won't have any of those. So the only answer choice C (17) is left which might be a factor of  16!+1 .

Answer: C.

ankitranjan · Answer

Bunuel You are the best.But there is one theorem that is Wilson's theorem...It states that 
 If n is a prime number ,(n-1)!+1 is divisible by n.

Hence 16!+1 i.e (17-1)! + 1 will be divisible by 17.

Bunuel · Answer

Yes, Wilson's theorem works for this particular number. Though you won't need this theorem for GMAT.

stonecold · Answer

Using the logic => Multiple +non multiple = non multiple => discarding all the options => only 17 is the viable choose..
Remember => we dont really have to compute the values here.. All is in the logic

bazu · Answer

Bunuel could you advise me if the following is a correct solution or if I am inventing math?

16! = 17!/17 --> 17!/17 + 1 =(17!+17)/17 ---> 16!+1 divisible by 17

Bunuel · Answer

16!+1=(17!+17)/17 but how do you conclude that (17!+17)/17 is divisible by 17? In other words how do you know that  /17 is an integer?

bazu · Answer

oh, yes I see, what I had in my mind does't actually make sense! thanks!

Mbawarrior01 · Answer

Learnt a new concept.
But when I apply the same to smaller factorials such as 5!+1 = 121 (divisible by 11 and not by 6-because of 3 & 2 in 5! but what about 7).

Is there any exception to the theorem?

broall · Answer

You could see this theorem here: https://en.wikipedia.org/wiki/Wilson's_theorem The theorem is applied for prime p and (p-1)!+1 divisible by p. Hence, 4!+1 is divisible by 5. However, 5!+1 will not be divisible by 6. This theorem could be proved by using advanced mathematic tools, thus this theorem is too hard and we no need to learn this theorem in solving GMAT PS/DS questions.

cavana · Answer

1) Two consecutive integers do not have any factors in common other than 1. This means that factor of 16!+1 has to be a number that is not a factor of 16!
2) 7, 1, and 6 are factors of 16! as they are multiples of 16!, 18 is also a factor of 16! because 2 and 9 are factors of 16!
3) By deduction, the only factor that is not a multiple of 16! is 17, and this means that it is a multiple of 16!+1

The correct answer is C.

ScottTargetTestPrep · Answer

We need to determine which one of the numbers in the given answer choices divides into 16! + 1. In other words, which one is a factor of 16!+1. To determine this, we must recognize that 16! and 16! + 1 are consecutive integers, and consecutive integers will never share the same prime factors. Thus, 16! and 16! + 1 must have different prime factors.

However, rather than breaking 16! factorial into primes, we can look at the answer choices to determine which choice is not a factor of 16!. Since 16! = 16 x 15 x 14…5 x 4 x 3 x 2 x 1, we see that choices A, B, and D are factors of 16! Since 18 = 2 x 9, 2 and 9 are also factors of 16!. However, none of these numbers (6, 7, 11, and 18) will be a factor of 16! + 1, so the only number that can be a factor of 16! + 1 is 17.

Answer: C

RahulJain293 · Answer

Here is how I went about this...

Unit's digit of 16! is 0 and + 1 gives 1 as unit's digit  --> Hence this is an ODD Number.

Eliminate D & E  as both are EVEN Numbers and hence cannot divide without leaving a remainder.

For A,B & C (7,11,17)

We know 16! has divisors 1 to 16 (option A&B divides 16!).. Meaning, 16!+1 (which is coprime to 16!) will not have an A&B as their divisors and hence only C (17) divides the number 16!+1

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Find the number that divides 16!+1?

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