Question

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

• A. 8
• B. 2
• C. 4
• D. 6

Answer 1

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