Multiples of 3 and 5
Problem 1
"If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000."
Though I am a C# developer by trade, I think python will give us the cleanest, most readable solution for this problem. Fastest? Maybe, maybe not  we can test that later :cowboy:
My initial solution to this problem was
Pretty embarrassing, right? Oh well, we all make stupid mistakes sometimes (or all the time) :withstupid:
If you don't see the issue with the above code, let me explain. The question was "Find the sum of all the multiples of 3 or 5 below 1000." The above solution does that, sort of  but it counts some numbers twice, for example 15. 15 is a multiple of both 3 and 5. Code revision time:
Perfect. It works! But... we did this in python for a reason. What's the point of using python if you don't even make the code pythonic? So, we can actually change this to be a pretty little one liner:
Aaaaaand I'm sure there is something even cooler, but I'm satisfied with this solution.
Put that in your interpreter, (make sure you understand the solution), then head over to Project Euler and save your answer for problem 1!
Thanks for reading.
Problem 1
"If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000."
Though I am a C# developer by trade, I think python will give us the cleanest, most readable solution for this problem. Fastest? Maybe, maybe not  we can test that later :cowboy:
My initial solution to this problem was
sum = 0 for x in range(1000): if x % 3 == 0: sum += x if x % 5 == 0: sum += x print(sum)
Pretty embarrassing, right? Oh well, we all make stupid mistakes sometimes (or all the time) :withstupid:
If you don't see the issue with the above code, let me explain. The question was "Find the sum of all the multiples of 3 or 5 below 1000." The above solution does that, sort of  but it counts some numbers twice, for example 15. 15 is a multiple of both 3 and 5. Code revision time:
sum = 0 for x in range(1000): if x % 3 == 0 or x % 5 == 0: sum += x print(sum)
Perfect. It works! But... we did this in python for a reason. What's the point of using python if you don't even make the code pythonic? So, we can actually change this to be a pretty little one liner:
print(sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0))
Aaaaaand I'm sure there is something even cooler, but I'm satisfied with this solution.
Put that in your interpreter, (make sure you understand the solution), then head over to Project Euler and save your answer for problem 1!
Thanks for reading.
1 Comments On This Entry
Page 1 of 1
MentalFloss
20 November 2018  06:08 PMprint(sum(set(range(0,1000,5))set(range(0,1000,3))))
print (the sum of (the set containing values 0 to 999 in increments of 5) union with (the set containing values 0 to 999 in increments of 3)).
Page 1 of 1
Trackbacks for this entry [ Trackback URL ]
← February 2019 →
S  M  T  W  T  F  S 

1  2  
3  4  5  6  7  8  9 
10  11  12  13  14  15  16 
17  18  19  20  21  22  23 
24  25  26  27  28 
Tags
My Blog Links
Recent Entries

Problem 1  The Beginning.
on Jan 03 2018 10:30 AM
Recent Comments
Search My Blog
1 user(s) viewing
1 Guests
0 member(s)
0 anonymous member(s)
0 member(s)
0 anonymous member(s)