If we list every the natural numbers below 10 that room multiples that 3 or 5, we obtain 3, 5, 6 and 9. The amount of this multiples is 23.

You are watching: Sum of multiples of 3 and 5 below 1000

Find the sum of all the multiples the 3 or 5 listed below 1000.


*

*

The previously posted answer isn"t correct. The explain of the difficulty is to amount the multiples the 3 and 5 listed below 1000, no up to and equal 1000. The exactly answer isegineqnarraysum_k_1 = 1^333 3k_1 + sum_k_2 = 1^199 5 k_2 - sum_k_3 =1^66 15 k_3 = 166833 + 99500 - 33165 = 233168,endeqnarraywhere we have the provided the identityegineqnarraysum_k = 1^n k = frac12 n(n+1).endeqnarray


*

$egingroup$ The one that posted the prize 233168, please define that price in detail . It would certainly be really helpful for us if you will explain. $endgroup$
First that all, prevent thinking ~ above the number $1000$ and also turn your attention to the number $990$ instead. If you deal with the trouble for $990$ friend just have to add $993, 995, 996$ & $999$ come it for the final answer. This sum is $(a)=3983$

Count all the #s divisible through $3$: native $3$... Come $990$ there are $330$ terms. The sum is $330(990+3)/2$, for this reason $(b)=163845$

Count every the #s divisible by $5$: native $5$... Come $990$ there space $198$ terms. The amount is $198(990+5)/2$, therefore $(c)=98505$

Now, the GCD (greatest usual divisor) that $3$ & $5$ is $1$, therefore the LCM (least typical multiple) have to be $3 imes 5 = 15$.

This way every number the divides through $15$ was counted twice, and also it must be done just once. Due to the fact that of this, you have an extra collection of numbers began with $15$ all the method to $990$ that needs to be gotten rid of from (b)&(c).

Then, from $15$... To $990$ there room $66$ terms and their sum is $66(990+15)/2$, therefore $(d)=33165$

The answer for the problem is: $(a)+(b)+(c)-(d) = 233168$

Simple but really fun problem.


re-superstructure
mention
follow
edited Feb 22 "14 in ~ 11:55
*

hardyellowcomic.com
34.5k1919 yellow badges6666 silver badges128128 bronze title
answer Feb 22 "14 at 9:13
*

agilaagila
31133 silver badges22 bronze title
$endgroup$
2
include a comment |
15
$egingroup$
The multiples the 3 space 3,6,9,12,15,18,21,24,27,30,....

The multiples the 5 space 5,10,15,20,25,30,35,40,45,....

The intersection that these two sequences is 15,30,45,...

The amount of the an initial numbers 1+2+3+4+...+n is n(n+1)/2.

The amount of the first couple of multiples of k, to speak k+2k+3k+4k+...+nk have to be kn(n+1)/2.

Now you can just put these ingredients with each other to deal with the problem.

Since we space asked come look for numbers below 1000, us shall look in ~ numbers up to the number 999.

To find n, usage 999/3 = 333 + remainder, 999/5 = 199 + remainder, 999/15 = 66 + remainder, by using a*(m*(m+1)/2) , whereby m=n/a. Here a is 3 or 5 or 15, and n is 999 or 1000 however 999 is best, and also then amount multiples the 3: $3((333)*(333+1)/2) = 166833$ add to multiples of 5: $5((199)*(199+1)/2) = 99500$; and also subtract multiples the 15 $15((66)+(66+1)/2 )= 33165$ to gain 233168.


share
point out
monitor
edited january 28 "20 in ~ 5:50
Mahesh muttinti
322 bronze title
reply Nov 7 "10 in ~ 12:36
anonanon
$endgroup$
2
add a comment |
9
$egingroup$
Well the key equation is already given above. The just question which provide me trouble is that why I have to subtract the sum of 15?! Well, the prize is, 15 can be evenly division by both 3 & 5. For this reason the commodities of 15 can also be split by those number as well! So, when you adding the number with sum Of 3 & sum Of five there space some numbers(i.e. 15,30,45,60....) i m sorry are available at both SUMMATION. So, you need to subtract at least once native the total sum to get the answer!

Hope this help someone favor me:) !!


share
mention
follow
answered Dec 28 "13 in ~ 11:22
enamenam
25922 silver badges77 bronze title
$endgroup$
add a comment |
Highly energetic question. Earn 10 reputation (not counting the combination bonus) in order to answer this question. The reputation necessity helps protect this inquiry from spam and also non-answer activity.

Not the price you're feather for? Browse other questions tagged number-theory project-euler or ask your own question.


Upcoming occasions
Featured ~ above Meta
Linked
7
What is the sum of every the organic numbers between $500$ and also $1000$.

See more: How Long Is A Block In Nyc, How Long Is A Standard City Block


4
Find the sum of every the multiples that 3 or 5 below 1000 (Break down)
connected
9
when do the multiples of 2 primes expectations all large enough natural numbers?
13
find the sum of the digits in the number 100!
1
CS problem, turned to yellowcomic.comematics
4
discover the amount of all the multiples that 3 or 5 listed below 1000 (Break down)
0
possible mis-interpretation in project Euler #21
3
Minimum amount of integers no multiples the one another.
2
recognize the sum of 3 numbers v 5 positive integer divisors.
hot Network inquiries more hot inquiries
inquiry feed
subscribe to RSS
question feed To subscribe to this RSS feed, copy and also paste this URL into your RSS reader.


yellowcomic.com
company
ridge Exchange Network
site design / logo © 2021 ridge Exchange Inc; user contributions licensed under cc by-sa. Rev2021.11.12.40742


yellowcomic.comematics ridge Exchange works best with JavaScript allowed
*

your privacy

By clicking “Accept all cookies”, girlfriend agree stack Exchange have the right to store cookie on your machine and disclose details in accordance through our Cookie Policy.