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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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