Question

find the units place as [2006]^2005 - [2005]^2006

Answer 1

The digit in the units place of $(2006^{2005}-2005^{2006})$ is $1$.

Step-by-step explanation:

The first number is $2006^{2005}$.

• We know that, $6^{1}=6$, $6^{2}=36$, $6^{3}=216$ and so on.

• So any positive integral power of $6$ ends with $6$ and thus we can say that the last digit of $2006^{2005}$ is $6$.

The second number is $2005^{2006}$.

• We know that, $5^{1}=5$, $5^{2}=25$, $5^{3}=125$ and so on.

• So any positive integral power of $5$ ends with $5$ and thus we can say that the last digit of $2005^{2006}$ is $5$.

Then, $2006^{2005}-2005^{2006}$

$=...6-...5$

$=...1$

The number in the units place is $1$.