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How to solve this problem, I can not figure it out:

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.

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    $\begingroup$ The question title does not seem to coincide with the question. For the given question: Take the sum of all multiples of 3, add the sum of all mutliples of 5 and subtract the sum of all multiples of 15. Use Gauß to calculate each sum. $\endgroup$ – Jendrik Stelzner Dec 31 '15 at 11:03
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Expanding on Jendrik's comment:

The sum of all multiples of $3$ below $1000$ is $S_3=3+6+...+999$. Notice that when we divide $S_3$ by $3$ we obtain $$ \frac{S_3}{3}=1+2+...+333 $$ which is the sum of all natural numbers between $1$ and $n_3=333$. Using Gauss, we know $$ 1+2+...+333=\frac{n_3(n_3+1)}{2} $$ which in this case equals $$ \frac{n_3(n_3+1)}{2}=\frac{333\cdot 334}{2}=333\cdot 167 $$ in other words, $$ \frac{S_3}{3}=333\cdot 167 $$ which implies $$ S_3=999\cdot 167 $$

The sum of all multiples of $5$ below $1000$ is $S_5=5+10+...+995$. By a very similar argument, we deduce $$ S_5=995\cdot 100 $$

However, if we add right away $S_3$ and $S_5$, we are adding twice each multiple of $15$ below $1000$. Therefore we must substract them, to obtain the result you are looking for.

The sum of all multiples of $15$ below $1000$ is $S_{15}=15+30+...+990$. By a very similar argument, we deduce $$ S_{15}=495\cdot 67 $$

The result is $S_3+S_5-S_{15}$.

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There is a method to find the multiple of 3: you should add all the digit and if it is also a multiple of 3, your number is multiple of three:

ex: 561 5+6+1=12 is multiple of 3 so 561 also

455 4+5+5=14 is not multiple of 3 so 455 also not

For the multiple of 5: the unit digit should be 0 or 5

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See it will be an ap of difference $3,5$ so its $Ap_1=3+6+...+999$ while for $5$ its $AP_2=5+10+15...995$ . But now you will have to subtract mutiples of $15$ as they ate counted twice so its another $AP_3$ which is $15+30...990$ thats it . Now you have fornula for summation of ap which is $\frac{n}{2}(2a+(n-1)d)$ where a is first term d is common difference,n is number of terms you can calculate number of terms from general term of ap which is $t_n=a+(n-1)d$ plug in values get the answer hope its clear.

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