Question

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

what class are you doing this in?

Answer 2

See, find the sum of all the multiples of 3 first that are coming under 1000:

That is:

3 + 6 + 9+............................. +999

This is An Arithmetic Progression having a = 3 and d = 3
Last term is :
999 = a + (n - 1)d

999 = 3 + 3n - 3

3n = 999
n = 333

So you can find the sum by using :

\[\huge Sum = \frac{n}{2}(a + l)\]

Where l = Last term = 999

So,

\[Sum = \frac{333}{2}(3 + 999) = 333 \times 501 = 166833\]

Now doing the same procedure for:

5 + 10 + 15+ ........................ + 995

a = 5, d= 5 l= 995

So,
\[Sum = \frac{199}{2}(5 + 995) = 199 \times 500 = 99500\]

Adding the two sums: 166833 + 99500 = 266333

Answer 3

There is one another method which is quite simpler then this..
If you want to know then tell I will explain it to you...

Answer 4

thnx for the reply! and yes @waterineyes i would also like to know it

Answer 5

See, you have to find first for:

3 + 6 + 9 + 12+ .................. + 999

Take 3 common:

3(1 + 2 + 3 + 4 + ......+333)

It is sum of first 333 natural numbers:

Its formula for sum is given by:

\[Sum = \frac{n(n+1)}{2}\]

Here n= 333 put and tell me what will be the sum..

Answer 6

55611

Answer 7

So, this is value of brackets;

so total sum = 3(55611) = ???

Answer 8

166833

Answer 9

Can you see by first method we have also the same result??

Answer 10

yes thnx very much!

Answer 11

Now solve for 5 + 10 + 15 + 20 ................+ 995

take 5 common and do the same steps as we have just done in finding the sum of 3 multiples..

Answer 12

ok thnx very much!

Answer 13

Welcome dear..