Stuck with factorials, some tips would be great!

[qwertz123]

Excuse my ingnorance, but how can you conclude the following?

"Ok, so now we know that 17!+1 is definitely prime,.."

Why is 17!+1 prime. I just don't get it

[D JAIN]

hi..can anyone pls tell me which book to refer for probability and permutation and combination....

[lime]

Excuse my ingnorance, but how can you conclude the following?
"Ok, so now we know that 17!+1 is definitely prime,.."

This is not your ignorance! This is the ignorance of the author of solution. Your question is absolutely valid!

17!+1 still might be divisible by 19, 23, 29 etc.

According to the google.com, it is the number of 10^14 order. I don't know any simple test to check it for being prime. So, I might only assume, that original question was saying that:

17!+1 < x <= 17!+17.

[sonikamadala]

Originally Posted by sunset_chill

Dear all,

I am totally puzzled about how to approach the following problems, can anybody help??
1.If (13!)^16 - (13!)^8/(13!)^8 + (13!)^4 = a , what is the unit’s digit of a/(13!)^4?
a)0
b)1
c)3
d)5
e)9




Thank you very much in advance!
sunset_chill

Imo answer is E.

Note : unit digit of 13! is 0 and a^2 - b^2 = (a+b)(a-b)

say 13! as x then equation becomes

a= (x)^16 - (x)^8 / (x)^8 + (x)^4 = { ((x)^8)^2 - ((x)^4)^2 } / (x)^8 + (x)^4
= { ((x)^8 + (x)^4) ((x)^8 - (x)^4) } / (x)^8 + (x)^4 = (x)^8 - (x)^4

a/(x)^4 = (x)^4 - 1

so here (x)^4 = (13!)^4 , unit digit for this is 0. so 0-1 is 9.

so answer should be B.

[gerald]

I have shorter way to solve the first question and I would like to share it with all of you.

First of all you should keep in mind nex formula: a^2-b^2= (a-b)(a+b)

Then, numerator can be expressed as (13!)^16-(13!)8 = ((13!)^8-(13!)^4)((13!)^8+(13!)^4). So we can cancel ((13!)^8+(13!)^4) in the numerator and denominator.
Then we are left with following expression (13!)^8-(13!)^4=a, As we know we should find a/(13!)^4. In other words we should divide (13!)^8-(13!)^4/ (13!)*4, Again after cancelation we are left with (13!)^2-1. The unit digit of (13!) is 0. So 0-1=9. The same answer.