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
Count the votes, counting each sophomore ballot as 1.5 votes and each freshmen ballot as 1 vote.
ur doing this because there is 200 more freshmen then sophomores...and if u count each sophomore vote as 1.5, it would make up for the 200 more freshmen
Answer:
No image or answer choices
Step-by-step explanation:
The new coordinate for C would be 1,2.
