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Question Number 165099 by SANOGO last updated on 26/Jan/22

$$therestofthedivisioneuclidienneof$$$${10}^{99}by13\times 17is?$$

Commented by Rasheed.Sindhi last updated on 26/Jan/22

$$\mathcal{T}ranslateinto\mathcal{E}nglishalso.$$

Answered by mr W last updated on 27/Jan/22

$$13\times 17=221$$$${10}^{99}={\left(1000\right)}^{33}$$$$={(4\times 221+116)}^{33}$$$$\models {116}^{33}$$$$=116\times {(60\times 221+196)}^{16}$$$$\models 116\times {196}^{16}$$$$=116\times {(173\times 221+183)}^8$$$$\models 116\times {183}^8$$$$=116\times {(151\times 221+118)}^4$$$$\models 116\times {118}^4$$$$=116\times {(60\times 221+664)}^2$$$$\models 116\times {664}^2$$$$=116\times (1995\times 221+1)$$$$\models 116=answer$$$$with\models imean$$$$‘‘hasthesameremainder{as}^{″}$$

Answered by Rasheed.Sindhi last updated on 27/Jan/22

$$\mathrm{Another}\mathrm{way}...$$$$Say,{10}^{99}\equiv x\left(mod221\right)[\because 13\times 17=221]$$$$\because \gcd (10,221)=1$$$$\therefore {10}^{\varphi \left(221\right)}\equiv 1\left(mod221\right)$$$$Now,\varphi \left(221\right)=221(1-\frac{1}{13})(1-\frac{1}{17})=192$$$$\therefore {10}^{192}\equiv 1\left(mod221\right)$$$$\mathcal{T}ryingfordicferentdivisorsof192$$$$Wecanseethat$$$${10}^{48}\equiv 1\left(mod221\right)$$$${\left({10}^{48}\right)}^2\equiv {\left(1\right)}^2\left(mod221\right)$$$${10}^{96}\equiv 1\left(mod221\right).........\left(i\right)$$$$Alsocanbeobservedthat$$$${10}^3\equiv 116\left(mod221\right)......\left(ii\right)$$$$\left(i\right)\times \left(ii\right):{10}^{99}\equiv 116\left(mod221\right)$$$$x=116$$

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