The last digit of the LCM of ( $3^{2003} - 1)$ and ( $3^{2003} + 1)$ is:

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Solution

The correct option is C 4
The given numbers are two consecutive even numbers, and therefore, their HCF = 2
Now: using LCM $\times$ HCF = product of two numbers
LCM $\times$ 2 = (...6) $\times$ (...8)
It can be seen now that the units digit of LCM = 4

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{\frac{3^{2003}}{28}}

{\frac{3^{2003}}{28}},

{.}

{\frac{3^{2003}}{28}}

\cdot

The value of ${\frac{3^{2003}}{28}}$ , where {.} denotes the fractional part, is equal to :

7^{3001} - 1

7^{3001} + 1.

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