Question

Write a python program to find the prime factorisation of an input number entered by the user.

Don't dare to spam, or I will report your account.

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

The given co‎de is written in Python.

n=b=int(input("Enter a number - "))

i=2

a=list()

while n!=1:

   if n%i==0:

       a.append(i)

       n//=i

   else:

       i+=1

print(b,"= ",end="")

print(*a,sep=" * ")

$\texttt{\textsf{\large{\underline{Logic}:}}}$

• Initialise i=2.
• Ask the user for the number (n).
• Loop while n!=1:
•   Check if n is divisible by i. If true, append the value of 'i' in the list.
•   If false, increment the value of i.
• At last, display the factors separated by ⋆.

Example: Consider a number, say 72.

>> n = 72

n != 1 is true.

>> n%i == 0 is true.

>> If block executes.

a = [2]

>> n = n/i

>> n = 36  [72/2]

>> n = 36

n !=1 is true.

>> n % i == 0 is true.

>> If block executes.

a = [2, 2]

>> n = n/i

>> n = 18 [36/2]

>> n = 18

n != 1 is true.

>> n % i == 0 is true.

>> If block executes.

a = [2, 2, 2]

>> n = n/i

>> n = 9 [18/2]

>> n = 9

>> n != 1 is true.

>> n % i  == 0 is false as 9 is not divisible by 2.

>> Else block executes.

>> i = i + 1

>> i = 3

>> n = 9

>> n !=1 is true.

>> n % i == 0 is true as 9 is divisible by 3.

>> If block executes.

a = [2, 2, 2, 3]

>> n = n/i

>> n = 3  [9/3]

>> n = 3

>> n !=1 is true.

>> n % i == 0 is true. as 3 is divisible by 3.

>> If block executes.

>>  a = [2, 2, 2, 3, 3]

>> n = n/i

>> n = 1  [3/3]

>> n = 1

>> n != 1 is false.

>> Loop is terminated.

So,

>> 72 = 2 * 2 * 2 * 3 * 3

In this way, problem is solved.