Find the unit digit of the LCM of 73001–1 and 73001+1.

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Find the unit digit of the LCM of $7^{3001} - 1$ and $7^{3001} + 1.$

Answer: Option D
:
D

option d

$7^{3001} - 1$ and $7^{3001} + 1$ are consecutive even natural numbers

Hence, their $H C F = 2$

$L C M = \frac{product of numbers}{HCF} = \frac{(7^{8001})^{2} - 1}{2} = \frac{7^{6002} - 1}{2}$

$7^{6002}$ ends with 9 (use the concept of power cycle)

So $7^{6002} - 1$ ends in 8. We also know that as $7^{6002}$ is a perfect square of an odd number, its tens digit will be even. So the required unit digit is 4.

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