Question

Here it is

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.

Answer 1

@dan815

Answer 2

sum of multiples of 3 + sum of multiples of 5 - sum of multiples of 15

Answer 3

lol

Answer 4

each of them is an arithmetic series

Answer 5

i did this already

Answer 6

u add the sums of 3 and sums of 5 and subtract the sums of 15

Answer 7

@hartnn dan gave me a link to solve math problem and I failed at the first question (this one) he laughed at me, hehehe

Answer 8

there are 333 terms that 3 can produce below 1000
and there are 199 terms 5 can produve below 1000
and there are 1000/15 -1 terms produced by 15

Answer 9

i can show u my code but its a little uugglaayy

Answer 10

how can you know there are 333 terms of 3?

Answer 11

1000/3

Answer 12

333*3=999 is the last term

Answer 13

1000/5 =200
1000 is the lst term and total 200 terms

Answer 14

oooh, there is a method !!!

Answer 15

*below*
so 995 last term and 199 terms

Answer 16

hartn come join us!! on our programming journey

Answer 17

thre's always a method xD

Answer 18

i'm doing enough programming for y work. :P no more programming while i enjoy my hobby :)

Answer 19

so, it is 330 + 199 -5 ??

Answer 20

whats that ?

Answer 21

oh 333 +190-5

Answer 22

you're counting no. of terms? why ??

Answer 23

you said so

Answer 24

take each series as separate arithmetic series

Answer 25

series 1
3+6+9+..999

a1 = 3, d = 3, n= 333
sum =...

Answer 26

http://www.compileonline.com/execute_python_online.php

Answer 27

y=[]
x=[]
for i in range(334):
a=3*i
y.append(a)

for i in range(200):
b=5*i
x.append(b)

for i in range(67):
x[3*i]=0
ans=sum(x)

print ans

Answer 28

series 2
5+10+15+..995

a1= 5,d=5, n = 199
sum2 =...

Answer 29

paste that code on that site and click excecute

Answer 30

see if it works

Answer 31

yes, I got it, thanks @hartnn

Answer 32

hartnn do u know python

Answer 33

did u get why do we do this ?
sum of multiples of 3 + sum of multiples of 5 - sum of multiples of 15

Answer 34

yes, I got the idea. hartnn

Answer 35

yes, i've learned python like a month ago, and i use it in my work

Answer 36

yha loser do u get that

Answer 37

you are adding that sum twice because
once for 5 * k3
and 3*k5

Answer 38

why for 15 you have to -1 term?

Answer 39

are you using my code

Answer 40

download python its free!

Answer 41

15|1000= 66.6666
so, the last term is 975., still <1000, why you have to -1?

Answer 42

its the easiest language to do some quick alogorythms

Answer 43

I don't know python. dan. I use calculator

Answer 44

lol

Answer 45

o mg plz dont!

Answer 46

you will get so stuck! this site teaches you some neat math after you solve it will show you more efficient ways

Answer 47

why??? when I studiedcomputer, I tried to understand the way the computer works, I asked and my prof hid himself to me, hahahaha...,

Answer 48

hahaha

Answer 49

look at this question! okay

Answer 50

OK, I will try python, but now, explain me. why you -1 from 15

Answer 51

what u mean

Answer 52

download python so u can run my code and see

Answer 53

15|1000 -1 term

Answer 54

i use 66

Answer 55

15*66 = 990

Answer 56

yes, Sir

Answer 57

so x=[]
for i in range(67): % this range 67 makes numbers go only to 67-1 = 66
a=15*i
x.append(a) % keeps adding every multiply of 15 to the list x

Answer 58

another way is to just ssum as you get them

Answer 59

sum1=0
for i in range(67):
sum1=sum1+15*i

Answer 60

sum1 will contain the addition of all the multiles of 15* 1 to 15*66

Answer 61

sum1=0
for i in range(200):
sum1=sum1+5*i

sum2=0
for i in range(334):
sum1=sum1+3*i

sum3=0
for i in range(67):
sum1=sum1+15*i

ans = sum1+sum2-sum3

Answer 62

this is a another complete CODE i just wrote up for u :) ENJOOOYy

Answer 63

got you, dan

Answer 64

sum1=0
for i in range(200):
sum1=sum1+5*i

sum2=0
for i in range(334):
sum2=sum2+3*i

sum3=0
for i in range(67):
sum3=sum3+15*i

ans = sum1+sum2-sum3

Answer 65

forgot to replace the numbers beside the sum :)

Answer 66

hey

Answer 67

do this question umm

Answer 68

lets derive the sum of 1^2+2^2+3^2....+100^2 with a formula

Answer 69

233168

Answer 70

no i want formula

Answer 71

1^2+2^2+...n^2 = ?

Answer 72

ohh u mean u solved the fisrt question ok

Answer 73

brb eating

Answer 74

this is answer for question 1 : 233168

Answer 75

me too. I go to "iit" something

Answer 76

look at question 7

Answer 77

no look at question 6

Answer 78

derive sum of square formula im eating