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Question Number 18135 by Tinkutara last updated on 15/Jul/17

$$\mathrm{Let}x\mathrm{be}\mathrm{the}\mathrm{LCM}\mathrm{of}{3}^{2002}-1\mathrm{and}$$$$3^{2002}+1.\mathrm{Find}\mathrm{the}\mathrm{last}\mathrm{digit}\mathrm{of}x.$$

Commented by mrW1 last updated on 15/Jul/17

$$0$$

Commented by prakash jain last updated on 16/Jul/17

$$\mathrm{If}\mathrm{A}\times \mathrm{B}\mathrm{is}\mathrm{divisible}\mathrm{by}10.\mathrm{Then}$$$$\mathrm{their}\mathrm{LCM}\mathrm{is}\mathrm{divisible}\mathrm{by}10.$$$$$$

Commented by mrW1 last updated on 16/Jul/17

$$3^1\mathrm{ends}\mathrm{with}3$$$$3^2\mathrm{ends}\mathrm{with}9$$$$3^3\mathrm{ends}\mathrm{with}7$$$$3^4\mathrm{ends}\mathrm{with}1$$$$\mathrm{generally}\mathrm{for}\mathrm{k}\in \mathrm{N}$$$$3^{4\mathrm{k}}\mathrm{ends}\mathrm{with}1$$$$3^{4\mathrm{k}+1}\mathrm{ends}\mathrm{with}3$$$$3^{4\mathrm{k}+2}\mathrm{ends}\mathrm{with}9$$$$3^{4\mathrm{k}+3}\mathrm{ends}\mathrm{with}7$$$$\mathrm{since}2002\equiv 2\left(\mathrm{mod}4\right)$$$$\Rightarrow 3^{2002}\mathrm{ends}\mathrm{with}9$$$$\Rightarrow 3^{2002}-1\mathrm{ends}\mathrm{with}8$$$$\Rightarrow 3^{2002}+1\mathrm{ends}\mathrm{with}0$$$$\Rightarrow \mathrm{a}\mathrm{multiple}\mathrm{of}{3}^{2002}+1\mathrm{ends}\mathrm{with}0$$$$\mathrm{since}\mathrm{LCM}\mathrm{of}{3}^{2002}-1\mathrm{and}{3}^{2002}+1\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}{3}^{2002}+1$$$$\Rightarrow \mathrm{LCM}\mathrm{of}{3}^{2002}-1\mathrm{and}{3}^{2002}+1\mathrm{ends}\mathrm{with}0$$

Commented by Tinkutara last updated on 16/Jul/17

$$\mathrm{But}\mathrm{why}\mathrm{do}\mathrm{you}\mathrm{multiply}\mathrm{the}\mathrm{numbers}$$$$\mathrm{for}\mathrm{their}\mathrm{LCM}?\mathrm{Their}\mathrm{HCF}\mathrm{is}\mathrm{not}1.$$

Commented by Tinkutara last updated on 16/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Commented by mrW1 last updated on 16/Jul/17

$$\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{multiply}\mathrm{the}\mathrm{both}\mathrm{numbers}\mathrm{but}$$$$\mathrm{say}\mathrm{that}\mathrm{the}\mathrm{LCM}\mathrm{of}{3}^{2002}-1\mathrm{and}{3}^{2002}+1$$$$\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}{3}^{2002}+1.\mathrm{since}\mathrm{every}$$$$\mathrm{multiple}\mathrm{of}{3}^{2002}+1\mathrm{ends}\mathrm{with}0,\mathrm{therefore}$$$$\mathrm{the}\mathrm{LCM}\mathrm{ends}\mathrm{also}\mathrm{with}0.$$$$\mathrm{in}\mathrm{this}\mathrm{case}{3}^{2002}-1\mathrm{could}\mathrm{be}\mathrm{every}\mathrm{other}$$$$\mathrm{number}\mathrm{p},\mathrm{the}\mathrm{result}\mathrm{is}\mathrm{the}\mathrm{same}:$$$$\mathrm{the}\mathrm{LCM}\mathrm{of}\mathrm{p}\mathrm{and}{3}^{2002}+1\mathrm{has}\mathrm{the}\mathrm{last}$$$$\mathrm{digit}0.$$

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