Your answer should be c if it is wrong try d. hope this helps :)
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:
The nth term is 109-9n
Step-by-step explanation:
Here, we want to find the nth term of the given arithmetic sequence
Mathematically, we have the nth term as;
Tn = a + (n-1)d
where a is the first term which is 100 in this case
d is the common difference which is the value obtained by subtracting the preceding term from the succeeding term; it is constant throughout the sequence
The value here is thus;
82-91 = 91-100 = -9
Substituting these values
Tn = 100 + (n-1)-9
Tn = 100 -9n + 9
Tn = 100 + 9 - 9n
Tn = 109-9n
2.74x7.5=20.55=20.550
That's all of your answers.
Answer:
5 dollars a week
Step-by-step explanation:
18-3=15
15/3=5
