If n = 1(1!) + 2(2!) + 3(3!) + 4(4!) + ... + 50(50!) and p is the smal

Question

If  n = 1(1!) + 2(2!) + 3(3!) + 4(4!) + ... + 50(50!)  and p is the smallest prime factor of n, then p is?

A. Between 2 and 20
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 50
E. Greater than 50

Are You Up For the Challenge: 700 Level Questions

nick1816 · Accepted Answer

n = 1(1!) + 2(2!) + 3(3!) + 4(4!) + ... + 50(50!)

x*x! = (x+1-1)*x! = (x+1)*x! - 1*x! = (x+1)! - x!

hence 1(1!) = 2!-1!
2(2!) = 3! - 2!

And so on

n=  2!  -  1!  +  3!  -  2!  +  4!  -  3!  +  5!  -  4!  + ........+  50!  -  49!  +  51!  -  50!

n = 51! - 1

51! is the multiple of all the prime numbers smaller than or equal to 47, but 1 is not. Hence, there difference can never be the multiple of any prime number that is smaller than or equal to 47.

Smallest prime factor of n is greater than 47 or 50 (there is no prime number between 47 and 50.

E

GMATinsight · Answer

Let, try this in steps

If n = 1*1! = 1 i.e. No prime divided it
If n = 1*1!+2*2! = 5 i.e. smallest prime factor = 5 which is greater than 2
If n = 1*1!+2*2!+3*3! = 23 i.e. smallest prime factor = 23 which is greater than 3
If n = 1*1!+2*2!+3*3!+4*4! = 119 i.e. smallest prime factor = 7 which is greater than 4
If n = 1*1!+2*2!+3*3!+4*4!+5*5! = 719 i.e. smallest prime factor = 719 which is greater than 5

similarly,  n = 1(1!) + 2(2!) + 3(3!) + 4(4!) + ... + 50(50!) , smallest prime factor must be GREATER than 50 hence

Answer: Option E

Avysar · Answer

We know that (n+1)n!= (n+1)!
So (n+1-1)n!= (n+1)n!!-n!=(n+1)!-n!
Using this for the given sequence
1(1!)+2(2!)+...50(50!)
we get
1(1!)=2!-1!
2(2!)=3!-2! and so on. 
So we have 2!-1!+3!-2!.....+51!-50!
Cancelling the positive and negative terms, we have 51!-1!
To solve this, let us understand from some smaller numbers
Let us see 3!-1. The smallest prime factor of 3! is 2 and 1 doesn't have a prime factor. And 3!-2=5. The smallest prime factor of 5 will be 5 which is >3
See 4!-1=23
Smallest prime factor of 4! Is 2 but no prime factor for 1 and the smallest prime factor of 4!-1=23 is 23 which is >4
Let us see 5!-1=119
Smallest prime factor of 5! is 2 and 1 doesn't have a prime factor..5!-1=119 and the smallest prime factor is 7 which is >5
So, proceeding further we can observe that as the number increases, the smallest prime factor for say x!-1 is greater than x.
Since the given sequence is 51!-1 the prime factor should be greater than 51...
Hence E

Posted from my mobile device

chetan2u · Answer

Each term is in the form of n(n!)=(n+1-1)n!=(n+1)n!-1*n!=(n+1)!-n!

So each term can be written as difference of factorial of one number higher and factorial of that number. 
1(1!)=2!-1!
2(2!)=3!-2!
3(3!)=4!-3!
......
49(49!)=50!-49!
50(50!)=51!-50!

Therefore, the expression 1(1!) + 2(2!) + 3(3!) + 4(4!)………….50(50!) can be written as
2!-1!+3!-2!+4!-3!+.......+50!-49!+51!-50!=51!-1!=51!-1

Now 51! is a multiple of all the numbers, including all the prime numbers less than 51. Hence 1 less than this number will not be a multiple of any number till 51.

So P>51>50

E

sahilvermani · Answer

Can an expert explain the above part?

chetan2u · Answer

Take a smaller number 
5!=1*2*3*4*5=120
So 120-1 or 119 will never be divisible by any of 2, 3, 4 or 5. Rather each of then will leave a remainder of -1.
So 5 will leave a remainder -1 or 5-1=4, while 4 will leave -1 or 4-1=3 and so on.

\frac{51!-1}{43}=\frac{51!}{43}-\frac{1}{43} 
51! will be divisible by 43 but -1 will remain.

abowlofrice · Answer

I see that n(n!)=(n+1-1)n! is required, but I would never think of doing this if I ever encountered this kind of problem. I guess my point is, what is the key takeaway from this problem?

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If n = 1(1!) + 2(2!) + 3(3!) + 4(4!) + ... + 50(50!) and p is the smal

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If n = 1(1!) + 2(2!) + 3(3!) + 4(4!) + ... + 50(50!) and p is the smal

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards