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
The answer is 2 x10 = 20
To the power of b which is 15
Answer:
the numbers are 20 and 24
Step-by-step explanation:
The numbers are in the ratio 5 : 6 = 5z : 6z , where z is a multiplier
The numbers sum to 44, thus
5z + 6z = 44
11z = 44 ( divide both sides by 11 )
z = 4
Hence
5z = 5 × 4 = 20
6z = 6 × 4 = 24
The two numbers are 20 and 24
Answer:
Step-by-step explanation:
The sum of an infinite geometric series is expressed according to the formula;
where;
a is the first term of the series
r is the common ratio
If the sum of an infinite geometric series is three times the first term, this is expressed as 
Substitute into the formula above to get the common ratio r;
open the parenthesis
subtract 3 from both sides
<em>Hence the common ratio of this series is </em><em></em>
