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Question Number 84941 by Power last updated on 17/Mar/20

Commented by mr W last updated on 17/Mar/20

$$usemethodinQ84782tofindthe$$$$lasttwodigits$$

Answered by mr W last updated on 17/Mar/20

$${348}^{67}$$$$\stackrel{2}{=}{48}^{67}=48\times {\left({48}^2\right)}^{33}$$$$\stackrel{2}{=}48\times 4^{33}=48\times 4\times {\left(4^4\right)}^8$$$$\stackrel{2}{=}48\times 4\times {56}^8=48\times 4\times {\left({56}^2\right)}^4$$$$\stackrel{2}{=}48\times 4\times {36}^4=48\times 4\times {\left({36}^2\right)}^2$$$$\stackrel{2}{=}48\times 4\times {96}^2$$$$\stackrel{2}{=}48\times 4\times 16=48\times 64$$$$\stackrel{2}{=}72$$$$\Rightarrow thelasttwodigitsof{348}^{67}are72$$$$\Rightarrow {348}^{67}\equiv 72mod\left(100\right)$$

Answered by mr W last updated on 18/Mar/20

$$whatifthequestionis$$$${348}^{67}\equiv xmod89$$$$$$$${348}^{67}={(3\times 89+81)}^{67}$$$$\equiv {81}^{67}mod89$$$$\equiv 81\times {(73\times 89+64)}^{33}mod89$$$$\equiv 81\times {64}^{33}mod89$$$$\equiv 81\times 64\times {(46\times 89+2)}^{16}mod89$$$$\equiv 81\times 64\times 2^{16}mod89$$$$\equiv 81\times 64\times {(2\times 89+78)}^{2}mod89$$$$\equiv 81\times 64\times {78}^{2}mod89$$$$\equiv 81\times 64\times (68\times 89+32)mod89$$$$\equiv 81\times 64\times 32mod89$$$$\equiv (58\times 89+22)\times 32mod89$$$$\equiv 22\times 32mod89$$$$\equiv (7\times 89+81)mod89$$$$\equiv 81mod89$$

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