Question Number 70969 by TawaTawa last updated on 10/Oct/19
Answered by mind is power last updated on 10/Oct/19
![first find how many zero ends 80! n=[((80)/5)]+[((80)/(25))]=16+3=19 so we have 80!=d∗10^(19) d∗10^(19) −x=0(10^(20) ) ⇒d∗10^(19) −x=10^(20) ∗r ⇒x∣10^(19) x=10^(19) .y ⇒d−y=10r ⇒d=y(10) d=((80!)/(10^(19) )) ((80!)/(10^(19) )) hase nofacteur of 5 he hase how many 2 m=[((80)/2)]+[((80)/4)]+[((80)/8)]+[((80)/(16))]+[((80)/(32))]+[((80)/(64))] =40+20+10+5+2+1=78 ((80!)/(10^(19) )) how too find it mod [10⌋ power of 2 is 2^(78−19) =2^(59) ((80!)/(10^(19) ))= 2^(59) .(all even number that are not multipl of 5) =2^(59) .(9.7.3.1)^8 mod(10) =2^(59) (−1.−3.3.1)^8 =2^(59) (10) 2^(59) mod(10) 2^1 =2(10),2^2 =4(10),2^3 =8(10) 2^4 =6(10),2^5 =2(10) ⇒2^(4k+3) =8(10) 59=4.14+3⇒2^(59) =8(10)⇒(((80)!)/(10^(19) ))=8(10) ⇒80!=8∗10^(19) +10^(20) .r ⇒80!=8∗10^(19) (10^(20) )](https://www.tinkutara.com/question/Q70973.png)
$${first}\:{find}\:{how}\:{many}\:{zero}\:{ends}\:\mathrm{80}! \\ $$$${n}=\left[\frac{\mathrm{80}}{\mathrm{5}}\right]+\left[\frac{\mathrm{80}}{\mathrm{25}}\right]=\mathrm{16}+\mathrm{3}=\mathrm{19} \\ $$$${so}\:{we}\:{have} \\ $$$$\mathrm{80}!={d}\ast\mathrm{10}^{\mathrm{19}} \\ $$$${d}\ast\mathrm{10}^{\mathrm{19}} −{x}=\mathrm{0}\left(\mathrm{10}^{\mathrm{20}} \right) \\ $$$$\Rightarrow{d}\ast\mathrm{10}^{\mathrm{19}} −{x}=\mathrm{10}^{\mathrm{20}} \ast{r} \\ $$$$\Rightarrow{x}\mid\mathrm{10}^{\mathrm{19}} \\ $$$${x}=\mathrm{10}^{\mathrm{19}} .{y} \\ $$$$\Rightarrow{d}−{y}=\mathrm{10}{r} \\ $$$$\Rightarrow{d}={y}\left(\mathrm{10}\right) \\ $$$${d}=\frac{\mathrm{80}!}{\mathrm{10}^{\mathrm{19}} }\:\:\: \\ $$$$\frac{\mathrm{80}!}{\mathrm{10}^{\mathrm{19}} }\:\:\:{hase}\:{nofacteur}\:{of}\:\mathrm{5}\:{he}\:{hase}\:{how}\:{many}\:\mathrm{2} \\ $$$${m}=\left[\frac{\mathrm{80}}{\mathrm{2}}\right]+\left[\frac{\mathrm{80}}{\mathrm{4}}\right]+\left[\frac{\mathrm{80}}{\mathrm{8}}\right]+\left[\frac{\mathrm{80}}{\mathrm{16}}\right]+\left[\frac{\mathrm{80}}{\mathrm{32}}\right]+\left[\frac{\mathrm{80}}{\mathrm{64}}\right] \\ $$$$=\mathrm{40}+\mathrm{20}+\mathrm{10}+\mathrm{5}+\mathrm{2}+\mathrm{1}=\mathrm{78} \\ $$$$\frac{\mathrm{80}!}{\mathrm{10}^{\mathrm{19}} }\:\:{how}\:{too}\:{find}\:{it}\:{mod}\:\left[\mathrm{10}\rfloor\right. \\ $$$${power}\:{of}\:\mathrm{2}\:{is}\:\mathrm{2}^{\mathrm{78}−\mathrm{19}} =\mathrm{2}^{\mathrm{59}} \\ $$$$\frac{\mathrm{80}!}{\mathrm{10}^{\mathrm{19}} }=\:\mathrm{2}^{\mathrm{59}} .\left({all}\:{even}\:{number}\:{that}\:{are}\:{not}\:{multipl}\:{of}\:\mathrm{5}\right) \\ $$$$=\mathrm{2}^{\mathrm{59}} .\left(\mathrm{9}.\mathrm{7}.\mathrm{3}.\mathrm{1}\right)^{\mathrm{8}} {mod}\left(\mathrm{10}\right) \\ $$$$=\mathrm{2}^{\mathrm{59}} \left(−\mathrm{1}.−\mathrm{3}.\mathrm{3}.\mathrm{1}\right)^{\mathrm{8}} =\mathrm{2}^{\mathrm{59}} \left(\mathrm{10}\right) \\ $$$$\mathrm{2}^{\mathrm{59}} {mod}\left(\mathrm{10}\right) \\ $$$$\mathrm{2}^{\mathrm{1}} =\mathrm{2}\left(\mathrm{10}\right),\mathrm{2}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{10}\right),\mathrm{2}^{\mathrm{3}} =\mathrm{8}\left(\mathrm{10}\right) \\ $$$$\mathrm{2}^{\mathrm{4}} =\mathrm{6}\left(\mathrm{10}\right),\mathrm{2}^{\mathrm{5}} =\mathrm{2}\left(\mathrm{10}\right) \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{4}{k}+\mathrm{3}} =\mathrm{8}\left(\mathrm{10}\right) \\ $$$$\mathrm{59}=\mathrm{4}.\mathrm{14}+\mathrm{3}\Rightarrow\mathrm{2}^{\mathrm{59}} =\mathrm{8}\left(\mathrm{10}\right)\Rightarrow\frac{\left(\mathrm{80}\right)!}{\mathrm{10}^{\mathrm{19}} }=\mathrm{8}\left(\mathrm{10}\right) \\ $$$$\Rightarrow\mathrm{80}!=\mathrm{8}\ast\mathrm{10}^{\mathrm{19}} +\mathrm{10}^{\mathrm{20}} .{r} \\ $$$$\Rightarrow\mathrm{80}!=\mathrm{8}\ast\mathrm{10}^{\mathrm{19}} \left(\mathrm{10}^{\mathrm{20}} \right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
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
$${withe}\:{pleasur}\:{Sir} \\ $$