Answer is X=2
Merry Christmas
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
Answer:
i wish i could hep i will do the math right now
Step-by-step explanation:
8*2=16, 4*4=16, so 16/2 is even and so is 16/8 or 16/4
♥ Solve:
(Arjun) 40*30=1200
(Dalia) 55*30=1650
Now subtract
1650-1200=450.
That means that Dalia will type 450 minutes more.
Final answer: 450
