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Question Number 13606 by Tinkutara last updated on 21/May/17

$$\mathrm{Show}\mathrm{that}{19}^{93}-{13}^{99}\mathrm{is}\mathrm{a}\mathrm{positive}$$$$\mathrm{integer}\mathrm{divisible}\mathrm{by}162.$$

Answered by mrW1 last updated on 21/May/17

$$\log \left({19}^{93}\right)=93\times \log 19\approx 119$$$$\log \left({13}^{99}\right)=99\times \log 13\approx 110$$$$\Rightarrow {19}^{93}>{13}^{99}$$$$$$$${19}^{93}={(1+18)}^{93}=1+93\times 18+\frac{93\times 92}{2!}\times {18}^2+\frac{93\times 92\times 91}{3!}\times {18}^3+...$$$$=A_0+A_1+A_2+A_3+...+A_{93}$$$$since{18}^2=324=2\times 162$$$$\Rightarrow A_{k⩾2}mod162=0$$$$A_0+A_1=1+93\times 18=1675=10\times 162+55$$$$\Rightarrow {19}^{93}mod162=55$$$$$$$${13}^{99}={(1+12)}^{99}=1+99\times 12+\frac{99\times 98}{2!}\times {12}^2+\frac{99\times 98\times 97}{3!}\times {12}^3+...$$$$=B_0+B_1+B_3+B_4+...+B_{99}$$$$since{12}^4=20736=128\times 162$$$$\Rightarrow B_{k⩾4}mod162=0$$$$3\times {12}^3=5184=32\times 162$$$$\Rightarrow B_{3}mod162=0$$$$9\times {12}^2=1296=8\times 162$$$$\Rightarrow B_{2}mod162=0$$$$$$$$B_0+B_1=1+99\times 12=1189=7\times 162+55$$$$\Rightarrow {13}^{99}mod162=55$$$$$$$$\Rightarrow ({19}^{93}-{13}^{99})mod162=55-55=0$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/May/17

$$sobeautiful.$$

Commented by RasheedSindhi last updated on 22/May/17

$$\mathrm{Great}!$$

Commented by Tinkutara last updated on 22/May/17

$$\mathrm{Thanks}\mathrm{but}\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{calculated}$$$$93\log 19\mathrm{and}99\log 13?\mathrm{In}\mathrm{the}\mathrm{textbook}$$$$\mathrm{it}\mathrm{was}\mathrm{mentioned}\mathrm{that}\mathrm{use}\mathrm{of}$$$$\mathrm{calculators}\mathrm{is}\mathrm{not}\mathrm{allowed}.$$$$\mathrm{Second},\mathrm{can}\mathrm{we}\mathrm{use}\mathrm{in}\mathrm{general}\mathrm{this}$$$$\mathrm{property}?$$$${(1+x)}^n\left(\mathrm{mod}y\right)=(1+nx)\left(\mathrm{mod}y\right)$$$$\mathrm{I}\mathrm{found}\mathrm{it}\mathrm{from}\mathrm{internet}\mathrm{from}\mathrm{Quora}$$$$\mathrm{by}\mathrm{typing}\mathrm{the}\mathrm{same}\mathrm{question}.$$

Commented by RasheedSindhi last updated on 22/May/17

$$\mathrm{Mentioned}\mathrm{logs}\mathrm{can}\mathrm{be}\mathrm{calculated}$$$$\mathrm{by}\mathrm{table}\mathrm{of}\mathrm{logs}\mathrm{instead}\mathrm{of}\mathrm{calculator}.$$$$$$

Commented by Tinkutara last updated on 22/May/17

$$\mathrm{But}\mathrm{it}\mathrm{was}\mathrm{to}\mathrm{be}\mathrm{proved}\mathrm{without}\mathrm{tables}.$$$$\mathrm{I}\mathrm{have}\mathrm{done}\mathrm{it}\mathrm{without}\mathrm{tables}\mathrm{as}\mathrm{follows}:$$$${\left(\frac{19}{13}\right)}^2=\frac{361}{169}>2\Rightarrow {\left(\frac{19}{13}\right)}^8>2^4>13$$$$\Rightarrow {19}^8>{13}^9$$$$\Rightarrow {19}^{88}>{13}^{99}$$$$\Rightarrow {19}^{93}>{13}^{99}$$$$\mathrm{The}\mathrm{only}\mathrm{thing}:\mathrm{Can}\mathrm{we}\mathrm{use}$$$${(1+x)}^n\left(\mathrm{mod}y\right)=(1+nx)\left(\mathrm{mod}y\right)?$$$$\mathrm{What}\mathrm{are}\mathrm{the}\mathrm{conditions}\mathrm{for}x,y\mathrm{and}n?$$

Commented by prakash jain last updated on 22/May/17

$$\mathrm{The}\mathrm{first}\mathrm{step}\mathrm{is}\mathrm{not}\mathrm{essential}\mathrm{for}$$$$\mathrm{proof}.$$$$\mathrm{Rest}\mathrm{of}\mathrm{the}\mathrm{proof}\mathrm{remains}\mathrm{valid}$$$$\mathrm{even}\mathrm{if}\mathrm{the}\mathrm{first}\mathrm{step}\mathrm{is}\mathrm{omitted}.$$

Commented by prakash jain last updated on 22/May/17

$$\mathrm{The}\mathrm{expression}\mathrm{need}\mathrm{not}\mathrm{be}+\mathrm{ve}$$$$\mathrm{for}\mathrm{calculating}\mathrm{mod}.$$$$-11\equiv 0\equiv -22\left(\mathrm{mod}11\right)$$

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