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Question Number 70969 by TawaTawa last updated on 10/Oct/19

Answered by mind is power last updated on 10/Oct/19

$$firstfindhowmanyzeroends80!$$$$n=\left[\frac{80}{5}\right]+\left[\frac{80}{25}\right]=16+3=19$$$$sowehave$$$$80!=d\ast {10}^{19}$$$$d\ast {10}^{19}-x=0\left({10}^{20}\right)$$$$\Rightarrow d\ast {10}^{19}-x={10}^{20}\ast r$$$$\Rightarrow x\mid {10}^{19}$$$$x={10}^{19}.y$$$$\Rightarrow d-y=10r$$$$\Rightarrow d=y\left(10\right)$$$$d=\frac{80!}{{10}^{19}}$$$$\frac{80!}{{10}^{19}}hasenofacteurof5hehasehowmany2$$$$m=\left[\frac{80}{2}\right]+\left[\frac{80}{4}\right]+\left[\frac{80}{8}\right]+\left[\frac{80}{16}\right]+\left[\frac{80}{32}\right]+\left[\frac{80}{64}\right]$$$$=40+20+10+5+2+1=78$$$$\frac{80!}{{10}^{19}}howtoofinditmod[10\rfloor$$$$powerof2is{2}^{78-19}=2^{59}$$$$\frac{80!}{{10}^{19}}=2^{59}.\left(allevennumberthatarenotmultiplof5\right)$$$$=2^{59}.{(9.7.3.1)}^{8}mod\left(10\right)$$$$=2^{59}{(-1.-3.3.1)}^8=2^{59}\left(10\right)$$$$2^{59}mod\left(10\right)$$$$2^1=2\left(10\right),2^2=4\left(10\right),2^3=8\left(10\right)$$$$2^4=6\left(10\right),2^5=2\left(10\right)$$$$\Rightarrow 2^{4k+3}=8\left(10\right)$$$$59=4.14+3\Rightarrow 2^{59}=8\left(10\right)\Rightarrow \frac{\left(80\right)!}{{10}^{19}}=8\left(10\right)$$$$\Rightarrow 80!=8\ast {10}^{19}+{10}^{20}.r$$$$\Rightarrow 80!=8\ast {10}^{19}\left({10}^{20}\right)$$$$$$$$$$$$$$$$$$

Commented by TawaTawa last updated on 10/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{thanks}\mathrm{for}\mathrm{your}\mathrm{time}$$

Commented by mind is power last updated on 10/Oct/19

$$withepleasurSir$$

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