1 .Which of the following Bare Bones programs is self-terminating Assume X > 0 A
ID: 3826554 • Letter: 1
Question
1.Which of the following Bare Bones programs is self-terminating Assume X > 0
A. while X not 0:
B. while X not 0:
decr X
C. decr X
while X not 0:
2: If c = x mod n then c is guaranteed to be in which of the following sets
a.{ 1, 2, 3, … n-1 }
b.{ 0, 1, 2, 3, …. n-1 }
c.{ -(n-1), -(n-2), -(n-3), … -2, -1 0 }
d.{ 1, 2, 3, …. n }
3. If xa mod n = c and xb mod n = d then xa+b mod equals
A. c + d mod n
B. c*d mod n
C. cd mod n
D. None of the above
4. Using the private keys n = 133 and d = 5, decrypt the message 7
a. 7
b. 37
c. 49
d. None of the above
Explanation / Answer
1. Which of the following Bare Bones programs is self-terminating Assume X > 0
A. while X not 0:
This runs in an infinite loop, if there is no loop control statement inside the loop.
B. while X not 0:
decr X
This will run the loop for a fixed number of times provided X is positive.
So, this is what your answer is.
C. decr X
while X not 0:
This also does pretty similar to the previous option.
So, this is also your answer choice.
2. If c = x mod n then c is guaranteed to be in which of the following sets.
When x is divided by n, the remainder is what c is.
The remainder will always be in the range 0 - n-1 in this case.
So, the answer is:
b. { 0, 1, 2, 3, …. n-1 }
3. If xa mod n = c and xb mod n = d then xa + b mod n equals:
xa mod n = c. So c lies in the range {0, n-1}.
xb mod n = d. So d also lies in the range {0, n-1}.
Now, xa + b mod ???
This is incomplete...
4. Using the private keys n = 133, and d = 5, decrypt the message 7.
n = 133 = 7 * 19.
d = 5.
P = 7^5 mod 133
= 16807 mod 133
= 49.
So, the answer is: c. 49.
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