Question 957196
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 2812.
<pre>
2812 divided by 3 is 937.333333...

So there are 937 multiples of 3 less than 2812,
the last of which is (937)(3)=2811

Their sum is

3+6+9+12+15+18+21+24+27+30+...+2805+2808+2811

Therefore we use the sum formula

{{{S[n]=expr(n/2)(a[1]+a[n])}}}, with n=937, a<sub>1</sub>=3, d=3.

{{{S[937]=expr(937/2)(3+2811)=expr(937/2)(2814)=1318359}}}

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2812 divided by 5 is 562.4

So there are 562 multiples of 5 less than 2812,
the last of which is (562)(5)=2810

Their sum is

5+10+15+20+25+30+35+40+45+...+2805+2810

Therefore we use the sum formula

{{{S[n]=expr(n/2)(a[1]+a[n])}}}, with n=562, a<sub>1</sub>=5, d=5.

{{{S[562]=expr(562/2)(5+2810)=(281)(2815)=791015}}}

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If we add those two sums together we get

1318359+791015 = 2109374

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However that's too much because it adds the integers which are
multiples of both 3 and 5 twice.  Those are the multiples of 15.
So we must find the sum of the multiples of 15 below 2812 and 
subtract that from the 2109374:

--------------

2812 divided by 15 is 187.46666...

So there are 187 multiples of 15 less than 2812,
the last of which is (187)(15)=2805

Their sum is

15+30+45+60+75+90+105+120+135+...2790+2805

Therefore we use the sum formula

{{{S[n]=expr(n/2)(a[1]+a[n])}}}, with n=187, a<sub>1</sub>=15, a<sub>187</sub>=2805 

{{{S[187]=expr(187/2)(15+2805)=expr(187/2)(2820)=263670}}}

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So subtracting 263670 from 2109374, the final answer is:

2109374-263670 = 1845704

Edwin</pre>