Answer:
Check the explanation
Step-by-step explanation:
(a)Let p be the smallest prime divisor of (n!)^2+1 if p<=n then p|n! Hence p can not divide (n!)^2+1. Hence p>n
(b) (n!)^2=-1 mod p now by format theorem (n!)^(p-1)= 1 mod p ( as p doesn't divide (n!)^2)
Hence (-1)^(p-1)/2= 1 mod p hence [ as p-1/2 is an integer] and hence( p-1)/2 is even number hence p is of the form 4k+1
(C) now let p be the largest prime of the form 4k+1 consider x= (p!)^2+1 . Let q be the smallest prime dividing x . By the previous exercises q> p and q is also of the form 4k+1 hence contradiction. Hence P_1 is infinite
I think it would be 64/100 because 64 percent is most likely out of 100 percent
Answer:5 percent
Step-by-step explanation: subtract new amount minus old amount then divide then make the decimal you get into and change to a percent
2x + 100 = 5x + 55
3x = 45
x = 15
2(15) + 100 = 130
180-130 = 50
so angle 2 is 50 degrees
