Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 174581 by Gbenga last updated on 04/Aug/22

$$3^{3^{3^3}}$$$$\mathbf{find}\mathbf{last}\mathbf{two}\mathbf{digits}$$$$$$$$$$$$$$$$$$

Answered by floor(10²Eta[1]) last updated on 05/Aug/22

$$3^{3^{3^3}}\left(\mathrm{mod}100\right)$$$$\phi \left(100\right)=\phi \left(5^2\right)\phi \left(2^2\right)=20.2=40$$$$\Rightarrow {\mathrm{a}}^{40}\equiv 1\left(\mathrm{mod}100\right),\gcd (\mathrm{a},100)=1$$$$\mathrm{note}\mathrm{that}:$$$$3^{27}={\left(3^4\right)}^6{3}^3={81}^6{3}^3\equiv 3^3=27\left(\mathrm{mod}40\right)$$$$\Rightarrow 3^{27}=40\mathrm{q}+27,\mathrm{for}\mathrm{some}\mathrm{q}\in \mathbb{Z}$$$$3^{3^{3^3}}=3^{3^{27}}=3^{40\mathrm{q}+27}={\left(3^{\mathrm{q}}\right)}^{40}{3}^{27}\equiv 3^{27}$$$$={81}^6{3}^3\equiv {19}^6{3}^3\left(\mathrm{mod}100\right)$$$$\mathrm{but}{19}^2=361\equiv -39\left(\mathrm{mod}100\right)$$$$\Rightarrow {19}^6{3}^3\equiv {39}^3{3}^3={117}^3\equiv {17}^3\left(\mathrm{mod}100\right)$$$$\mathrm{but}{17}^2=289\equiv -11\left(\mathrm{mod}100\right)$$$$\Rightarrow {17}^3\equiv -11.17=-187\equiv 13\left(\mathrm{mod}100\right)$$$$\Rightarrow 3^{3^{3^3}}\equiv 13\left(\mathrm{mod}100\right)$$$$\mathrm{so}\mathrm{the}\mathrm{last}\mathrm{two}\mathrm{digits}\mathrm{are}1\mathrm{and}3$$

Commented by Gbenga last updated on 14/Aug/22

$$\mathbf{thankw}\mathbf{wk}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com