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Let-x-be-the-LCM-of-3-2002-1-and-3-2002-1-Find-the-last-digit-of-x-




Question Number 18135 by Tinkutara last updated on 15/Jul/17
Let x be the LCM of 3^(2002)  − 1 and  3^(2002)  + 1. Find the last digit of x.
$$\mathrm{Let}\:{x}\:\mathrm{be}\:\mathrm{the}\:\mathrm{LCM}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} \:−\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{3}^{\mathrm{2002}} \:+\:\mathrm{1}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{of}\:{x}. \\ $$
Commented by mrW1 last updated on 15/Jul/17
0
$$\mathrm{0} \\ $$
Commented by prakash jain last updated on 16/Jul/17
If A×B is divisible by 10. Then  their LCM is divisible by 10.
$$\mathrm{If}\:\mathrm{A}×\mathrm{B}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{10}.\:\mathrm{Then} \\ $$$$\mathrm{their}\:\mathrm{LCM}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{10}. \\ $$$$ \\ $$
Commented by mrW1 last updated on 16/Jul/17
3^1  ends with 3  3^2  ends with 9  3^3  ends with 7  3^4  ends with 1  generally for k∈N  3^(4k)  ends with 1  3^(4k+1)  ends with 3  3^(4k+2)  ends with 9  3^(4k+3)  ends with 7  since 2002≡2 (mod 4)  ⇒3^(2002)  ends with 9  ⇒3^(2002) −1 ends with 8  ⇒3^(2002) +1 ends with 0  ⇒a multiple of 3^(2002) +1 ends with 0  since LCM of 3^(2002) −1 and 3^(2002) +1 is a multiple of 3^(2002) +1  ⇒LCM of 3^(2002) −1 and 3^(2002) +1 ends with 0
$$\mathrm{3}^{\mathrm{1}} \:\mathrm{ends}\:\mathrm{with}\:\mathrm{3} \\ $$$$\mathrm{3}^{\mathrm{2}} \:\mathrm{ends}\:\mathrm{with}\:\mathrm{9} \\ $$$$\mathrm{3}^{\mathrm{3}} \:\mathrm{ends}\:\mathrm{with}\:\mathrm{7} \\ $$$$\mathrm{3}^{\mathrm{4}} \:\mathrm{ends}\:\mathrm{with}\:\mathrm{1} \\ $$$$\mathrm{generally}\:\mathrm{for}\:\mathrm{k}\in\mathrm{N} \\ $$$$\mathrm{3}^{\mathrm{4k}} \:\mathrm{ends}\:\mathrm{with}\:\mathrm{1} \\ $$$$\mathrm{3}^{\mathrm{4k}+\mathrm{1}} \:\mathrm{ends}\:\mathrm{with}\:\mathrm{3} \\ $$$$\mathrm{3}^{\mathrm{4k}+\mathrm{2}} \:\mathrm{ends}\:\mathrm{with}\:\mathrm{9} \\ $$$$\mathrm{3}^{\mathrm{4k}+\mathrm{3}} \:\mathrm{ends}\:\mathrm{with}\:\mathrm{7} \\ $$$$\mathrm{since}\:\mathrm{2002}\equiv\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{2002}} \:\mathrm{ends}\:\mathrm{with}\:\mathrm{9} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{2002}} −\mathrm{1}\:\mathrm{ends}\:\mathrm{with}\:\mathrm{8} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{2002}} +\mathrm{1}\:\mathrm{ends}\:\mathrm{with}\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} +\mathrm{1}\:\mathrm{ends}\:\mathrm{with}\:\mathrm{0} \\ $$$$\mathrm{since}\:\mathrm{LCM}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} −\mathrm{1}\:\mathrm{and}\:\mathrm{3}^{\mathrm{2002}} +\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} +\mathrm{1} \\ $$$$\Rightarrow\mathrm{LCM}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} −\mathrm{1}\:\mathrm{and}\:\mathrm{3}^{\mathrm{2002}} +\mathrm{1}\:\mathrm{ends}\:\mathrm{with}\:\mathrm{0} \\ $$
Commented by Tinkutara last updated on 16/Jul/17
But why do you multiply the numbers  for their LCM? Their HCF is not 1.
$$\mathrm{But}\:\mathrm{why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{multiply}\:\mathrm{the}\:\mathrm{numbers} \\ $$$$\mathrm{for}\:\mathrm{their}\:\mathrm{LCM}?\:\mathrm{Their}\:\mathrm{HCF}\:\mathrm{is}\:\mathrm{not}\:\mathrm{1}. \\ $$
Commented by Tinkutara last updated on 16/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by mrW1 last updated on 16/Jul/17
I didn′t multiply the both numbers but  say that the LCM of 3^(2002) −1 and 3^(2002) +1  is a multiple of 3^(2002) +1. since every  multiple of 3^(2002) +1 ends with 0, therefore  the LCM ends also with 0.  in this case 3^(2002) −1 could be every other  number p, the result is the same:  the LCM of p and 3^(2002) +1 has the last  digit 0.
$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{multiply}\:\mathrm{the}\:\mathrm{both}\:\mathrm{numbers}\:\mathrm{but} \\ $$$$\mathrm{say}\:\mathrm{that}\:\mathrm{the}\:\mathrm{LCM}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} −\mathrm{1}\:\mathrm{and}\:\mathrm{3}^{\mathrm{2002}} +\mathrm{1} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} +\mathrm{1}.\:\mathrm{since}\:\mathrm{every} \\ $$$$\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} +\mathrm{1}\:\mathrm{ends}\:\mathrm{with}\:\mathrm{0},\:\mathrm{therefore} \\ $$$$\mathrm{the}\:\mathrm{LCM}\:\mathrm{ends}\:\mathrm{also}\:\mathrm{with}\:\mathrm{0}. \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{3}^{\mathrm{2002}} −\mathrm{1}\:\mathrm{could}\:\mathrm{be}\:\mathrm{every}\:\mathrm{other} \\ $$$$\mathrm{number}\:\mathrm{p},\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}: \\ $$$$\mathrm{the}\:\mathrm{LCM}\:\mathrm{of}\:\mathrm{p}\:\mathrm{and}\:\mathrm{3}^{\mathrm{2002}} +\mathrm{1}\:\mathrm{has}\:\mathrm{the}\:\mathrm{last} \\ $$$$\mathrm{digit}\:\mathrm{0}. \\ $$

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