All u do is divide 3 into 74 but 3 cant get into 74 so u do the number the close in that 3 times 24 in that 72 but if u do 25 times 3 that 75 in the number cant be greater so then u know it 24 times 3 then u subtract 74 from 72 in get 2 as the remainder so u have 2 as the hold number in 3 as the nemonator in 24 as the denatorator
There is two hundred and four (204) on a standard chessboard.
Is it −26−(6)^2? If so
−26−(6)^2
= -26 - 36
= - 62
If −26−(6)2 (means 6 times 2) then
= -26 - 12
= -38
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
