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Calculate the sum of all three-digit natural numbers divisible by five.

Correct answer:

s =  98550

Step-by-step explanation:

$$\begin{gathered} n_1=100\mathrm{/}5=20 \\ {n}_2=\lfloor 999\mathrm{/}5\rfloor =199 \\ n=n_2-n_1+1=199-20+1=180 \\ {s}_1={\displaystyle \frac{n_1+n_2}{2}}\cdot n={\displaystyle \frac{20+199}{2}}\cdot 180=19710 \\ s=5\cdot s_1=5\cdot 19710=98550 \\ \text{ Verifying Solution: } \\ {s}_2=100+105+110+115+120+125+130+135+140+145+150+155+160+165+170+175+180+185+190+195+200+205+210+215+220+225+230+235+240+245+250+255+260+265+270+275+280+285+290+295+300+305+310+315+320+325+330+335+340+345+350+355+360+365+370+375+380+385+390+395+400+405+410+415+420+425+430+435+440+445+450+455+460+465+470+475+480+485+490+495+500+505+510+515+520+525+530+535+540+545+550+555+560+565+570+575+580+585+590+595+600+605+610+615+620+625+630+635+640+645+650+655+660+665+670+675+680+685+690+695+700+705+710+715+720+725+730+735+740+745+750+755+760+765+770+775+780+785+790+795+800+805+810+815+820+825+830+835+840+845+850+855+860+865+870+875+880+885+890+895+900+905+910+915+920+925+930+935+940+945+950+955+960+965+970+975+980+985+990+995=98550 \end{gathered}$$

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