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 .

Commented by mrW1 last updated on 15/Jul/17

0

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

If A \times B is divisible by 10 . Then

their LCM is divisible by 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 \in N

3^{4 k} ends with 1

3^{4 k + 1} ends with 3

3^{4 k + 2} ends with 9

3^{4 k + 3} ends with 7

since 2002 \equiv 2 (\mod 4)

\Rightarrow 3^{2002} ends with 9

\Rightarrow 3^{2002} - 1 ends with 8

\Rightarrow 3^{2002} + 1 ends with 0

\Rightarrow 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

\Rightarrow LCM of 3^{2002} - 1 and 3^{2002} + 1 ends with 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 .

Commented by Tinkutara last updated on 16/Jul/17

Thanks 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 .

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