Question

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

• A. 8
• B. 6
• C. 9
• D. 4
• E. 2

Answer 1

The correct option is D 4

option d

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

Hence, their $HCF=2$

$LCM=\frac{\text{product of numbers}}{\text{HCF}}=\frac{(7^{8001}{)}^2-1}{2}=\frac{7^{6002}-1}{2}$

$7^{6002}$ 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.