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.
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
, with n=937, a1=3, d=3.
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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
, with n=562, a1=5, d=5.
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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:
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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
, with n=187, a1=15, a187=2805
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So subtracting 263670 from 2109374, the final answer is:
2109374-263670 = 1845704
Edwin