Project Euler #1: Sum of Multiples of 3 and 5 below 1000 Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Project Euler Problem 12 in PythonFunction to return the sum of multiples within a rangeProject Euler 78: Coin Partitions in PythonProject Euler #14: Longest Collatz sequenceProject Euler #15 in PythonProject Euler #4 - Largest Palindrome Project - PythonProject Euler problem #1 in Python 3Sum of multiples of 3 or 5 using functional programmingProject Euler 001 in C++Project Euler Problem 37 in Python
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Project Euler #1: Sum of Multiples of 3 and 5 below 1000
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Project Euler Problem 12 in PythonFunction to return the sum of multiples within a rangeProject Euler 78: Coin Partitions in PythonProject Euler #14: Longest Collatz sequenceProject Euler #15 in PythonProject Euler #4 - Largest Palindrome Project - PythonProject Euler problem #1 in Python 3Sum of multiples of 3 or 5 using functional programmingProject Euler 001 in C++Project Euler Problem 37 in Python
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
def prob_1():
sum_mult=[] #Create an empty list which will take sum of multiples of 3 and 5
check_sum=0
for i in range(1,1000): #Take numbers till 1000
#if(i)
if( (i%3)==0 or (i%35)==0 ): #divisor condition
sum_mult.append(i)
return sum(sum_mult) #return sum of list
I am just starting out my journey as a programmer, here is my code and I would love to see any critical feedback and other alternative solutions maybe using some clever hack of using lambda function's etc.
python performance python-3.x programming-challenge
New contributor
$endgroup$
add a comment |
$begingroup$
def prob_1():
sum_mult=[] #Create an empty list which will take sum of multiples of 3 and 5
check_sum=0
for i in range(1,1000): #Take numbers till 1000
#if(i)
if( (i%3)==0 or (i%35)==0 ): #divisor condition
sum_mult.append(i)
return sum(sum_mult) #return sum of list
I am just starting out my journey as a programmer, here is my code and I would love to see any critical feedback and other alternative solutions maybe using some clever hack of using lambda function's etc.
python performance python-3.x programming-challenge
New contributor
$endgroup$
add a comment |
$begingroup$
def prob_1():
sum_mult=[] #Create an empty list which will take sum of multiples of 3 and 5
check_sum=0
for i in range(1,1000): #Take numbers till 1000
#if(i)
if( (i%3)==0 or (i%35)==0 ): #divisor condition
sum_mult.append(i)
return sum(sum_mult) #return sum of list
I am just starting out my journey as a programmer, here is my code and I would love to see any critical feedback and other alternative solutions maybe using some clever hack of using lambda function's etc.
python performance python-3.x programming-challenge
New contributor
$endgroup$
def prob_1():
sum_mult=[] #Create an empty list which will take sum of multiples of 3 and 5
check_sum=0
for i in range(1,1000): #Take numbers till 1000
#if(i)
if( (i%3)==0 or (i%35)==0 ): #divisor condition
sum_mult.append(i)
return sum(sum_mult) #return sum of list
I am just starting out my journey as a programmer, here is my code and I would love to see any critical feedback and other alternative solutions maybe using some clever hack of using lambda function's etc.
python performance python-3.x programming-challenge
python performance python-3.x programming-challenge
New contributor
New contributor
edited 44 secs ago
AJNeufeld
6,9641723
6,9641723
New contributor
asked 48 mins ago
DjVasuDjVasu
11
11
New contributor
New contributor
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I expect you made a typo. You don't want (i%35)==0
, you want (i%5)==0
.
check_sum
is unused, and can be omitted.
The brackets around the if( ... ):
condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings"""
to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...)
function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
$endgroup$
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
$begingroup$
I expect you made a typo. You don't want (i%35)==0
, you want (i%5)==0
.
check_sum
is unused, and can be omitted.
The brackets around the if( ... ):
condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings"""
to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...)
function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
$endgroup$
add a comment |
$begingroup$
I expect you made a typo. You don't want (i%35)==0
, you want (i%5)==0
.
check_sum
is unused, and can be omitted.
The brackets around the if( ... ):
condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings"""
to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...)
function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
$endgroup$
add a comment |
$begingroup$
I expect you made a typo. You don't want (i%35)==0
, you want (i%5)==0
.
check_sum
is unused, and can be omitted.
The brackets around the if( ... ):
condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings"""
to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...)
function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
$endgroup$
I expect you made a typo. You don't want (i%35)==0
, you want (i%5)==0
.
check_sum
is unused, and can be omitted.
The brackets around the if( ... ):
condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings"""
to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...)
function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
answered 2 mins ago
AJNeufeldAJNeufeld
6,9641723
6,9641723
add a comment |
add a comment |
DjVasu is a new contributor. Be nice, and check out our Code of Conduct.
DjVasu is a new contributor. Be nice, and check out our Code of Conduct.
DjVasu is a new contributor. Be nice, and check out our Code of Conduct.
DjVasu is a new contributor. Be nice, and check out our Code of Conduct.
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