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Question Number 35005 by Rasheed.Sindhi last updated on 14/May/18

$$\mathrm{Find}\mathrm{remainder}\mathrm{when}$$$$1!+2!+3!+...+99!+100!$$$$\mathrm{is}\mathrm{divided}\mathrm{by}12.$$

Answered by rahul 19 last updated on 14/May/18

$$9.$$$$yesdidthesamewayasdoneby$$$$mjssir.$$

Commented by Rasheed.Sindhi last updated on 14/May/18

$$\mathrm{Right}!$$

Commented by rahul 19 last updated on 14/May/18

☺️☺️

Answered by MJS last updated on 14/May/18

$$\mathrm{since}4!=24\Rightarrow \mathrm{remainder}\mathrm{of}(\underset{n=4}{\sum ^{\mathrm{\infty }}}n!)/12\mathrm{is}\mathrm{zero}$$$$\mathrm{so}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{looking}\mathrm{for}\mathrm{the}\mathrm{remainder}\mathrm{of}$$$$(1!+2!+3!)/12=9/12=9$$

Commented by Rasheed.Sindhi last updated on 14/May/18

†håñk§ §ïr!

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