## Python: Factorial Digit Sum

**Python: Factorial Digit Sum **

** Problem 20 @ projectEuler**

**The Task:** The task in projectEuler P20 is to get the sum of the digits in the number Factorial of 100!

**Factorial Ndefinition**The factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example, 10! = 10 × 9 × … × 3 × 2 × 1 = 3628800.

**Problem 20** is another easy problem in projectEuler, and we will write two functions to solve it. __First one__ is a Factorial_digit_sum this one will return the factorial of a number. __The second__ function will calculate the sum of all digits in a number N and we will call it sum_of_digits.

**Clarification** As long as i just start solving or posting my answers to projectEuler portal, i am selecting problems and not going through them in sequence, that’s way my posts are jumps between problems, so if i am posting the code to solve problem 144 *(for example)* that does’t meaning that i solve all problems before it.

print of solved screen:

**The Code:**

#Python: Factorial Digit Sum

#Problem No.20 on projectEuler

def Factorial_digit_sum(num):

if (num == 1) :

return 1

else:

return num * Factorial_digit_sum(num-1)

num=100

fact =Factorial_digit_sum(100)

print fact,’is the Factorial of {}.’.format(num)

def sum_of_digits(dig):

t = 0

for each in dig:

t = t + int(each)

print ‘\nThe sum of your number is’,t

sum_of_digits(str(tot1))

## Python: Collatz Sequence

**Python: Longest Collatz Sequence **

**Problem No.14 in ProjectEuler**

**Definition** Wikipedia: Start with any positive integer n. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half the previous term. If the previous term is odd, the next term is 3 times the previous term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

So the formula is: Starting with n, the next n will be:

n/2 (if n is even)

3n + 1 (if n is odd)

If we start with 13, we generate the following sequence:

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1.

**A walk through: **

n=13, 13 is odd, then n = 13 * 3 + 1 , n= 40

n=40, 40 is even, then n= 40/2, n=20

n=20, 20 is even, then n=20/2, n=10

n=10, 10 is even, then n=10/2, n=5

n=5, 5 is odd, then n=5*3+1 , n=16

n=16, 16 is even, then n=16/2, n=8

n=8, 8 is even, then n=8/2, n=4

n=4, 4 is even, then n=4/2, n=2

n=2, 2 is even, then n=2/2, n=1

n=1 then end of sequence

**The Task:** The task in ProjectEuler is to searching for a the Number N, under one million, that produces the longest chain.

Overview to my python cases:In my company, we are not allowed to download any software, so i don’t have any Python platform. To solve this i am using an online python interpreter, some time it’s become slow. So in this code (and others) i am spiting the range in 10 each with 100,000 then running the code to get the longest chain in each range. So the Number N, under one million, that produces the longest chain is:

**The Answer: **In my previous codes or math solving challenges in pybites or ProjectEuler I am solving the problems, writing the code, but not posting my answer to ProjectEuler platform. Today, and with Problem No.14 i decide to post the answer in the ProjectEuler platform for the first time just to see what will happen. The answer was 837799, and I get this page.

*In the code bellow, i set the range from 1 to 50000.*

**The Code:**

chain2=[]

longest=[0,0]

def collatz_Seq(num):

t= num

chain=[num]

while t !=1 :

if t%2==0:

t=t/2

chain.append(int(t))

else:

t=3*t+1

chain.append(int(t))

return chain

for num in range (1,50000):

chain2 = collatz_Seq(num)

if len(chain2) > longest[0]:

longest[0] = len(chain2)

longest[1] = num

chain2=[]

print(‘num:’,longest[1],’ has a longest chain: ‘,longest[0])

## Python: The Factorial

**Python: Factors of the Number N**

This is a short task to get the factors of a given number. The Definition of Factors of N is: The pairs of numbers you multiply to get the N number.

For instance, factors of 15 are 3 and 5, because 3×5 = 15. Some numbers have more than one factorization (more than one way of being factored). For instance, 12 can be factored as 1×12, 2×6, or 3×4

In this task we will write a Python code to ask the user for a number N then will get all the pairs number that if we multiply them will get that N number, we will store the pairs in a array ‘factors’.

**The Code:**

def factors_of_n(num):

a=1

factors=[]

while a <= num:

if num%a==0:

if (num/a,a) not in factors:

factors.append((a,int(num/a)))

a = a + 1

return factors

#Ask the user for a number

num=int(input(“Enter a number: “))

print(factors_of_n(num))

## Python: Amicable Numbers

**Python: Amicable Numbers **

** Problem No.21 on projectEuler **

In this task we need to calculate the SUM of all divisors of N, from 1 to N.

Then we calculate the Sum of (sum of N divisors ) let’s say M .

Now if

**For example**, the proper divisors of **A=220 **are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore F(220) = 284. And the proper divisors of **B=284 **are 1, 2, 4, 71 and 142; so F(284) = 220. So F(A) =B and F(B)=A then A and B are amicable.

Definition:If f(a) = b and f(b) = a, where a ≠ b,

thena and b are anamicable pairand each of a and b are calledamicable numbers.

**In this task** we will ask the user to enter a Range of numbers and we will search for all Amicable Numbers pairs in that range(from – to) if we fond one we will print out the number and the divisors list. The main function here is the one that get the sum of divisors, we will call it get_divisors_sum and we will examine Amicable with If statement.

Hint ..

To check if the sum of divisors from both said are equal we use (t2==x )

AND that this sum are not same we use (t1!=t2)

AND that we did not print the pair before we use: (t2 not in ami_pair)

**The Code:**

#Python: Amicable Numbers

#Problem No.21 on projectEuler

num1=int(input(‘The range Start from:’))

num2=int(input(‘The range Ends at:’))

t1=0

t2=0

l=[]

l1=[]

l2=[]

ami_pair=[]

def get_divisors_sum (num):

t=0

l=[]

for a in range (1,num):

if num%a==0 :

l.append(a)

t=t+a

return t,l

for x in range(num1,num2):

l1=[]

t1,l1=get_divisors_sum(x)

l2=[]

t2,l2=get_divisors_sum(t1)

if (t2==x) and (t1!=t2) and (t2 not in ami_pair):

print(‘\nget_divisors_sum({}) is {} divisors={}’.format(x,t1,l1))

print(‘\nget_divisors_sum({}) is {} divisors={}’.format(t1,t2,l2))

print(‘\nSo {} and {} are Amicable Numbers .’.format(t1,t2))

ami_pair.extend((t1,t2))

## Python Project: Sum of power of digits

**Python: Sum of power of digits **

** Problem No.16 **

In Problem No.16, projectEuler ask to find the Sum of power of digits in the number, for example if we have 2^15 (2 to power of 15) the answer is 23768 then we need to calculate the sum of this number (2+3+7+6+8 ) that’s equal to 26.

**The Task:** So our task in this project is to find the sum of the digits of (2^1000). To write this as a program and to make it more __general__ we will ask the user to input the number and the power he want, then I start thinking to restrict user from input large numbers that could cause CPU problems, but then i decide to keep it open as is.

**The Code:**

num=2

p=15

num=int(input(“Enter a number “))

p=int(input(“Enter a power “))

a=num**p

def sum_of_digits(num,p):

a=num**p

l=[int(i) for i in str(a)]

print(l)

for x in range (len(l)):

t=t+l[x]

print (“sum =”,t)

## Python: Prime Numbers in Range

**Python: Prime Numbers in a Range**

Our task here is simple as the title, we will have a range and will test each number to see if it is a prime then we will add it to a list, once we finish we will print out the list.

To complete this task we will use one of our function we create last time (**Read:** is prime post).

So, here we will ask the user to input two numbers num1 and num2 the we will pass all the numbers in the range to ** is_prime()** and store the result in a list.

**The Code:**

#Function to get all Prime numbers in a range.

#Ask the user to enter two numbers

print ‘Get all Prime numbers in a range\n’

num1=int(input(“Enter the first number in the range: “))

num2=int(input(“Enter the last number in the range: “))

#create the list

prime_list=[]

def get_all_prime(num1,num2):

for x in range (num1,num2):

if is_prime(x) ==’Prime’:

prime_list.append(x)

return prime_list

#Call the function and print the list

get_all_prime(num1,num2)

print prime_list

print len(prime_list)

## Python Project

**Python: Bite 107 **

** Filter numbers with a list comprehension**

I was happy to join **Pybites ** and start to solve the code-challenges, my problem from the first bite was that i am solving the challenges in other online python interpreter, and it looks working fine and the output seams as it should be, but once I copy the code in Pybite portal and doing the code test it fail and gives errors.

In this challenge **Bite 107**, after a list of numbers the task is to return the positive even number. Now in the task description it mentioned to use (__elegant list comprehension; one line of code__). I start to work on it, immediately it jump to my head a for loop and if each item is even and more than 0 add it to my_list, then return my_list. After coding, the output was a list with positive and even numbers in a Number-list, but it failed in the Bite test portal. WAY!.

This type of failure was against me in other challenges also, and i was just setting all the day thinking what happen!. With **Bite 107** I decide to click on “**Show Solution**” button just to see what is wrong.

**One line code:**The Solution was in one line of list brackets after a return statement congaing for and if without my_list. just like that. I start smiling, and ask my self “**How could i know that i can write it like this??”**. Yes it is beautiful, Yes elegant and yes one line of code, and yes my-way of coding gives the answer that we want, my code pass my testing exam.

**Concloguen**: I think i come up with a good resource of learning python, I don’t want to read book’s but reading codes that doing things and having the alternatives of writing same code; this will help me a lot. I am not coding for NASA or in other sensitive area that processing time is essential and micro-milli seconds make difference, so one line of four-line to-me is OK. 🙂

**Acknowledge**: I’d like to thank **Pybite **community for there contribution in spreading knowledge and codes, and for this challenges. I will read the challenges, and try my best to solve them as elegant as i can, also i will not stop in one point because my code is 2 lines more than what should be.

**Last thing**, Bite 107 here is my code and the elegant code. Be calm and positive.

**The Code:**

**#First: My Code**

def filter_positive_even_numbers(numbers):

my_list=[]

for i in range(len(numbers)):

if (i>0 and i%2==0) :my_list.append(i)

return my_list

print (filter_positive_even_numbers(numbers))

**#The optimal solution **

def filter_positive_even_numbers(numbers):

return[n for n in numbers if n > 0 and n % 2 == 0]