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what is the most efficient way of getting digit sum? [closed]

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I have to find the digit sum until its in single digit.
if input is -9999 then output should be -9
(-(9+9+9+9))==-9)
if input is 9012 then output should be 3 (+(9+0+1+2)==1+2==3))
ps: i have solved this but the negative inputs are giving wrong output.I am using the division by 9 property(you can google it) in order to get o(1) solution
my code:
def digSum(n):
if (n == 0):
return 0
if (n % 9 == 0):
return 9
else:
(n % 9)

just a simple recursion should help, add this to your code
if n < 0:
return - digSum(abs(n))

Your code is close, but you need to check to see if the input is a negative number BEFORE you evaluate if it modulates 9 or not. Your current code evaluates if n % 9 == 0 and returns 9, but you should add a condition to check if it's negative and n % 9 first, otherwise it will always return a positive number.
Modify your code to be this:
def digSum(n):
if (n == 0):
return 0
if (n % 9 == 0 and n < 0):
return -9
if (n % 9 == 0):
return 9
else:
(n % 9)

How to find the sum of all the divisors of a number, without including it [closed]

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Finding the sum of all the divisors of a number, without including that particular number that is being divided.
For example if we want to find the divisors of number 6 they would be 1,2,3. Number 6 should be included as it is a divisor, but for this scenario we are not considering it.
Note:- number 3 and 4 are not included since, when you divide number 6 by number 3, there would be a remainder.
A divisor is a number that divides into another without a remainder.
def sum_divisors(n):
sum = 0
# Return the sum of all divisors of n, not including n
return sum
print(sum_divisors(0)) # 0
print(sum_divisors(3)) # Should sum of 1 # 1
print(sum_divisors(36)) # Should sum of 1+2+3+4+6+9+12+18 # 55
print(sum_divisors(102)) # Should be sum of 2+3+6+17+34+51 # 114

I think I got it:
def sum_divisors(n):
return sum([i for i in range(1, n)
if n % i == 0])
print(sum_divisors(0))
print(sum_divisors(3)) # Should sum of 1
print(sum_divisors(36)) # Should sum of 1+2+3+4+6+9+12+18
print(sum_divisors(102)) # Should be sum of 2+3+6+17+34+51
Output
0
1
55
114

You could use a combined filter() and lambda() function
num = 36
div_sum = sum(list(filter(lambda div : num % div == 0 , range(1, num))))
# divisors : [1, 2, 3, 4, 6, 9, 12, 18]
# divisors sum : 55
print('divisors sum : ', div_sum)

We need to find the divisors of the number first, and then sum them. Finding out if a number is a divisor of another number is pretty simple:
if x % y == 0:
is_divisor = True
Modulo - % - returns the remainder of a division. 9 % 2 = 1. 10 % 3 = 1. However, we'll need to know which numbers to check. We could just loop through every number up to our input:
for i in range(n):
if n % i == 0:
is_divisor = True
...but that's too much. For example, we know that the highest possible divisor of n that isn't n itself, is 1/2n if n is even, and lower if it's not. Instead, let's focus on getting each pair, and then check if we should continue. We also need to store all of these numbers, so lets create some lists:
lower_divisors = []
upper_divisors = []
We'll want to know what our current highest lower divisor and lowest highest divisor are, and also know where to start. We can populate the lower divisor list with 1 and the highest with n. We can worry about removing n later.
lower_divisors = [1]
upper_divisors = [n]
continu = True
while continu:
# Let's get the last item in each list
highest_low = lower_divisors[-1]
lowest_high = upper_divisors[-1]
for i in range(highest_low + 1, lowest_high):
if n % i == 0:
lower_divisors.append(i)
upper_divisors.append(n // i)
break
continu = False
# If we've left the loop, we have all of our divisors.
lower_sum = sum(lower_divisors)
higher_sum = sum(upper_divisors) - n #Gotta take it back out!
sm = lower_sum + higher_sum
return sm
That should be a fast way to achieve your goal, without making your computer work harder - but... it's a lot more steps to write. I wrote it in far more steps to make it clear what I was doing and to be readable, but it could definitely be trimmed up a lot.
Upon running the code, the simple method of looping through each number from 1 to n starts to really add up when n is over 8 digits. 9 digit numbers were taking around 12-16 seconds on my machine. The longer approach above returns 32-digit values for n in under 1/10th of a second.

Python programming and the use of for loops [closed]

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Am new to programming and I just started studying and practicing. I started with python and am at for loops now. I kinda coded something that’s getting confusing for me because I can’t seem to understand how the code arrived at that output. Can someone please explain it to me. Would be very grateful.
Here’s the code I did:
a = range(1,20)
total = 0
for i in a :
if i%3==0 or i%5==0 :
total new = total + i
print (total new)
And the output was 18.

a = range(1,20)
Your program is creating a range of numbers between 1 and 19 (ranges go to one less than the max number you specify) like [1, 2, 3, 4, ..., 19]
total = 0
Your program is initializing the variable total to equal 0
for i in a:
You start looping through the range you made earlier, the first iteration i=1, the next i=2 and so on until i=19
if i % 3 == 0 or i % 5 == 0:
You are selecting only the data where the modulo of 5 or 3 is 0. For example:
3 % 3 == 0 (0 remainder)
4 % 3 == 1 (1 remainder)
Now altering your variable to a reasonable name (without spaces) that will actually utilize the variable we initialized above
total = total + i # alternatively written "total += i"
This says that every time the value i is evenly divisible by 3 or 5 we will add it to our total
print(total)
We show the final result after adding values. You incorrectly scripted your program to do this though so it only showed the largest value that was evenly divisible by 5 or 3 which is 18.
When scripted correctly:
a = range(1, 20)
total = 0
for i in a:
if i % 3 == 0 or i % 5 == 0:
total += i
print(total)
Outputs
78

Python Modularity for a number [duplicate]

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How do I sort a dictionary by value?
(34 answers)
Closed 5 years ago.
How to create a itterative number 10digit long by taking a base number

A Simple approach for Solution:
a=input()
while len(a)<10:
b=int(a[len(a)-1])+int(a[len(a)-2])
a=a+str(b%10)
print (a)
May This one helpful for you.

n= 16
while(len(str(n)) < 10):
print(n)
a = n %10 + (int(n/10))%10 # sum of last two digits
n *= 10
n += a%10 #adding the last
print(n)

Based on your example, I'm assuming you want to stop once you reach a 0.
Certain numbers, like 18 wouldn't reach a 0 ever, so also want to add a meximum length that the ID can be.
The following code runs the math until you reach a 0, or until it reaches 40 digits.
def getID(number)
maxlength = 40 - len(str(number))
while maxlength:
string = str(number)
answer = (int(string[-2]) + int(string[-1])) % 10
if answer == 0:
break
number = number * 10 + answer
maxlength -= 1
return number
getID(14) == 1459437
getID(15) == 15617853819
getID(16) == 1673
getID(18) == 17853819
getID(18) == 189763921347189763921347189763921347189763
18 is a number that repeats forever, so instead the loop ends once it reaches a set length.

Python:Project Euler #1 [closed]

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I've tried looking up some questions that were already on StackOverflow for project Euler problem 1 but not much has helped me. This is my code, it runs, but it's not correct. not sure whats wrong?
def sumOfMultiples(number):
sum = 0
for i in range(1,number):
if i % 3 == 0 or i % 5 == 0:
sum += 1
return sum
print (sumOfMultiples(15))

You are not using i at all. When the number meets the condition (divisible by 3 or 5), you want to add it, not 1.
sum += 1
should be
sum += i

The question asks for "the sum of all the multiples of 3 or 5 below 1000". You're adding 1 to the sum instead of adding the multiple of 3 or 5 to the sum.
So sum += 1 should be sum += i.

import time
#Recording start time of the program
start=time.time()
#A variable to store the sum
result=0
#calculating the sum for all the numbers below 1000
for i in range(1,1000):
#Checking the the condition
if i%3==0 or i%5==0:
result=result+i
print("Sum of all numbers that are multiples of 3 or 5 below 1000 is",result)
print("Execution time of program is",time.time()-start)

def sumOfMultiples(number):
sum = 0
for i in range(1,number):
if i % 3 == 0 or i % 5 == 0:
sum += i
return sum
print (sumOfMultiples(15))
In place of sum+=1 write sum+=i.

Here is a single-liner for you to ponder:
print sum([x for x in xrange(1,1000) if x%3==0 or x%5==0])
or you could also do it like this:
print sum([[0,x][x%3==0 or x%5==0] for x in xrange(1,10**3)])

try using this:
def sumOfMultiples(number):
sum = 0
for i in range(1,number):
if i % 3 == 0 or i % 5 == 0:
sum += i ## You needed to add i, not 1. Probably just a type-o as they are similar looking characters.
print(sum)
(sumOfMultiples(1000)) ## Also Euler1 requires this input to be 1000 for correct answer.