Suppose that (n, e) is an rsa encryption key, with n = pq, where p and q are large primes and gcd(e, (p − 1)(q − 1)) = 1. furthermore, suppose that d is an inverse of e modulo (p − 1)(q − 1). suppose that c ≡ me (mod pq). in the text we showed that rsa decryption, that is, the congruence cd ≡ m (mod pq) holds when gcd(m, pq) = 1. show that this decryption congruence also holds when gcd(m, pq) > 1. [hint: use congruences modulo p and modulo q and apply the chinese remainder theorem.]

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages

twenty-six miles over two gallons because the proportion for the original 13 miles to one gallon would be written 13/1 and, doing the same thing to the top and bottom, i multiplied by 2 and got 26/2

where does the z at is it on the x axes or on the y axes any way if it is on the y axes put it horizontally and if it is on the x axes put it vertically

step-by-step explanation: