Let n!=1*2*3*...*n for n>=1.If p=1!+(2*2!)+(3*3!)+...+(10*10!),then p+2 when divided by 11! leaves remainder?

Question

Let n!=1*2*3*...*n for n>=1.If p=1!+(2*2!)+(3*3!)+...+(10*10!),then p+2 when divided by 11! leaves remainder?
(a) 10
(b) 0
(c) 7
(d) 1

Answer 1

Do your own Maths homework boy!

Answer 2

Basically, we first have to evaluate p.
p = (1*1!) + (2*2!) + ... + (10*10!)
= (2! - 1!) + (3! - 2!) + ... + (11! - 10!)
= 11! - 1!
p + 2 = 11! - 1! + 2
= 11! + 1
Thus p + 2 would give a remainder of 1 (answer d) when divided by 11!.

Answer 3

enable' s remedy the 1st section first. we can discover the LCM of 5,6,8,9,12 and upload one million to the top result to remedy section one million. when you consider that 6 divides 12, that's adequate to come across the LCM of 5 8 9 and 12. when you consider that 8 and 9 are particularly top(i.e., have not have been given any top ingredient in common) their LCM is their product or seventy two. So we choose the LCM of 5,seventy two and 12. when you consider that 12 divides seventy two it fairly is the LCM of 5 and seventy two. ultimately, when you consider that 5 and seventy two are particularly top, their LCM is 360. So the answer to section one million is 361. Now we ought to discover the smallest ok such that 360k + one million is divisible by way of 13. by way of trial, we detect that ok = 10 and the respond is 3601. right this is yet in a different thank you to do the final section: we could like 360k +one million = 0(mod 13) 360 ok = 12(mod 13) 30k = one million(mod 13) 4k = one million(mod 13) so ok = 10(mod 13) and the respond is 3601. Sorry for my till now mistake!