If p = 1! + (2 × 2!) + (3 × 3!) + … + (11 × 11!), then p + 2 when

Question

If p = 1! + (2 × 2!) + (3 × 3!) + … + (11 × 11!), then p + 2 when divided by 12! Leaves a remainder of

A. 0
B. 1
C. 2
D. 8
E. 11

preetamsaha · Accepted Answer

n(n!) = (n+1 - 1)(n!) = (n+1!) - (n!)
p = 1! + (2 × 2!) + (3 × 3!) + … + (11 × 11!),
or, p= 1+ (3!) - (2!) +(4!) - (3!) + … + (12!) -(11!)
or, p= 1 - 2 + (12!)
or,p+2 = 1+ (12!)

so, p + 2 when divided by 12! Leaves a remainder of 1.

correct answer is B

CrackverbalGMAT · Answer

In questions like these, observing patterns by calculating the first few terms of the sequence is important. IN fact, obtaining the pattern is the cornerstone of solving any problem on sequences. 
When you see a question with a number of terms and the ellipsis symbols (the dots basically), know that the question is one on sequences. And at that instant, you should look at observing and developing a pattern.

Let’s look at evaluating the given expression. 
P = 1! + (2x2!) + (3x3!) +…. + (11x11!). The fact that there is a 11x11! in the expression should tell us that it’s pointless to evaluate each expression. And that’s one of the hallmarks of a question on sequences, there will be one or two terms that look intimidating individually, but fit in nicely into the pattern if you can find one.
1! = 1
2X2! = 2x2 = 4
3X3! = 3x6 = 18.

No clue at this stage? Observe the expression. A fairly huge expression with lots of big numbers to be added. What if we bring in some subtraction and cancel out some terms? It will reduce the number of additions we have to do.
1! = 2! – 1!
2X2! = 4 = 3! – 2!
3X3! = 18 = 4! – 3!.
This is a neat little pattern developing here. So now you understand that a lot of terms will cancel out.

So, p = 2! – 1! + 3!-2! + 4! – 3! + 5! – 4!+……. + 12! – 11!. When we cancel out terms, we see that we will be left with p = 12! – 1!.

Therefore, p+2 = 12! + 1. When 12! + 1 is divided by 12!, clearly, the remainder is 1.

The correct answer option is B .

Hope that helps!

zs2 · Answer

If p = 1! + (2 × 2!) + (3 × 3!) + … + (11 × 11!), then p + 2 when divided by 12! Leaves a remainder of

A. 0
B. 1
C. 2
D. 8
E. 11

For someone like me who is not able to find out patterns ( particularly under stress and pressure of an exam). You can always look for a basic process.... One thing is sure that GMAT does not want us to find out and solve such big question stems... There must be some shortcut or logic behind the question....

One think is that stem goes up to 11 * 11! and division is asked by 12! ( some logic must be here)

What if we divide ---
1! + (2 × 2!) by 3!

1! + (2 × 2!) + (3 × 3!) by 4!

1! + (2 × 2!) + (3 × 3!) + (4 × 4!) by 5! .....ans so on ...

In all above cases is that Remainder is always ( -1 )......

So p = 1! + (2 × 2!) + (3 × 3!) + … + (11 × 11!) by 12! will also give remainder (-1).....

If p gives remainder -1 , p+2 will give remainder -1+2 = 1

Hence B is the answer.....

Oppenheimer1945 · Answer

Using n n!=(n+1-1)n!=(n+1)!-n!

p=12!-1
p+2=12!+1

Remainder when p+2 is divided by 12! Is 1

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VGSH · Answer

This exp may be easy to some of you 
P= 1! + (2*2!) + (3*3!) +..........+ (11*11!)
1! = 1
2*2! = 4
3*3! = 18
4*4! = 96
from this we could see that from 4! every single number will be a multiple of 12, so it will leave a remainder of 0.
now, we'll take 1+4+18, which equals 23. Adding 2 to P makes it 25, and when 25 is divided by 12 it leaves a remainder of 1. 
so the ans is 1
Option B

VGSH · Answer

This exp may be easy to some of you 
P= 1! + (2*2!) + (3*3!) +..........+ (11*11!)
1! = 1
2*2! = 4
3*3! = 18
4*4! = 96
from this we could see that from 4! every single number will be a multiple of 12, so it will leave a remainder of 0.
now, we'll take 1+4+18, which equals 23. Adding 2 to P makes it 25, and when 25 is divided by 12 it leaves a remainder of 1. 
so the ans is 1
Option B

If p = 1! + (2 × 2!) + (3 × 3!) + … + (11 × 11!), then p + 2 when

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