q26

[Saumyojit]

Let n! = 1 x 2 x 3 x……….x n for integer n> 1. If p = 1! + (2 x 2!) + (3 x 3!) + ……(10 x 10!),then p+2 when divided by 11! Leaves remainder of
(a) 10
(b) 0
(c) 7
(d) 1

How to approach?

 

[Dr.Peterson]

Let n! = 1 x 2 x 3 x……….x n for integer n> 1. If p = 1! + (2 x 2!) + (3 x 3!) + ……(10 x 10!),then p+2 when divided by 11! Leaves remainder of
(a) 10
(b) 0
(c) 7
(d) 1

How to approach?

If you don't have any immediate ideas, I would try finding these sums for smaller numbers, and look for a pattern relating such a sum to the next factorial. You will see one, and then can think about why it might continue to be true.

Another approach starts by noting that every term in your sum is less than 11!, but the last is close. What if you subtract the last term from 11!, and then repeat?

 

[Saumyojit]

If you don't have any immediate ideas, I would try finding these sums for smaller numbers, and look for a pattern relating such a sum to the next factorial. You will see one, and then can think about why it might continue to be true.

Another approach starts by noting that every term in your sum is less than 11!, but the last is close. What if you subtract the last term from 11!, and then repeat?

I found one notation right now ,
n!=(n+1) ! - n * n!

But how will it be used in this sum?

 

[Dr.Peterson]

I found one notation right now ,
n!=(n+1) ! - n * n!

But how will it be used in this sum?

Did you "find" that by searching for what others have said, or by thinking, which is what you need to learn to do?

Just think. Don't practice dependency; practice intellectual independence by boldly going where you don't yet know you can go.

First, why is that equation true? (Did you try factoring (n+1)! as (n+1)*n!, which is a standard fact about factorials?) Don't just believe what you read; understand it. When you understand a tool, you can use it effectively.

Next, try doing various things with it. If n=11, what does it look like? Is that at all relevant to the problem?

Or work in the other direction, following my first suggestion. How does 1! + 2*2! relate to 3! ? What about 1! + 2*2! + 3*3! ?

 

[Saumyojit]

p= (2-1)!*1! + (3-1)*2! ....(10- 1)*9!= 10!- 1

p=10! - 1

This came?

 

[Saumyojit]

ok done

 

[Dr.Peterson]

p= (2-1)!*1! + (3-1)*2! ....(10- 1)*9!= 10!- 1

p=10! - 1

This came?

I'm not sure of your reasoning there, though it looks similar to something I might do. Can you explain more fully?

And what does your conclusion have to do with the question about dividing by 11!?