Answer:
- common = []
- num1 = 8
- num2 = 24
- for i in range(1, num1 + 1):
- if(num1 % i == 0 and num2 % i == 0):
- common.append(i)
- print(common)
Explanation:
The solution is written in Python 3.
Firstly create a common list to hold a list of the common factor between 8 and 24 (Line 1).
Create two variables num1, and num2 and set 8 and 24 as their values, respectively (Line 3 - 4).
Create a for loop to traverse through the number from 1 to 8 and use modulus operator to check if num1 and num2 are divisible by current i value. If so the remainder of both num1%i and num2%i will be zero and the if block will run to append the current i value to common list (Line 6-8).
After the loop, print the common list and we shall get [1, 2, 4, 8]
#accepting input from user
n=int(input("Enter a number: "))
#entered number is stored in a temporary variable
temp=n
#initializng required variables
rev=0
dgt=0
#digits are reversed inside while loop
while(n>0):
dgt=n%10
rev=rev*10+dgt
n=n//10
#original number and its reverse are compared
if(temp==rev):
#if equal, it's a palindrome
print("It is a Palindrome")
else:
#if not equal, it's not a palindrome
print("It is not a Palindrome")
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Answer:
Check the explanation
Explanation:
A) Whenever C is sending to D, what other communications are possible?
C’s packet will be seen by A, B and D, but not by E. Thus, D can send to E at the sametime..
B) Whenever B is sending to A, what other communications are likely?
Even though B’s packet will not be seen by D, other nodes, e.g., E, or C, can’t send to D since the packets from these nodes will interfere with the packets from B at A. Therefore, other communications is not likely at the same time.
C) Whenever B is sending to C, what other communications are possible?
B’s packet will be seen by E, A and C, by not by D. therefore, E can send to D at the same point.
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