Only one digit

Number Theory Level 3

What is the last digit of the lowest common multiple of $(3^{2002}+1)$ and $(3^{2002}-1)$ ?

The answer is 0.

1 solution

Md Zuhair
Mar 18, 2017

It is quite easy

The LCM must be the product of the two = $3^{4004}-1$

Hence

We can see by Euler's Totient Function

$\phi(10) = 4$

SO $3^4 = 1 \mod 10$

$3^{4004} = 1 \mod 10$ $\text{[Powers are rasied to 1001 ]}$

Great!, but just one input. GCD of the two numbers is 2, we can say this as both the numbers are necessarily even. So the LCM is half the product of the two numbers, not particularly the product of the two numbers.

Yash Jain - 4 years, 2 months ago

Yes, That is correct

Md Zuhair - 4 years, 2 months ago

How do you prove that the last digit is 0, not 5?

Freddie Hand - 4 years, 2 months ago

@Freddie Hand The tenth digit will be even as the pattern tells-

$3^3=27$

$3^4=81$

$3^5=243$

$3^6=729$

and so on $\cdots$

Yash Jain - 4 years, 2 months ago

Only one digit

The answer is 0.

1 solution

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