3. What is the 'tens' digit of $2013^{2}-2013$ ?
A 0
B 1
C 4
D 5
E 6
Solution
D The 'tens' digit of $2013^{2}-2013$ is the same as that of $13^{2}-13$. Since $13^{2}-13=169-13=$ 156 , the 'tens' digit of $2013^{2}-2013$ is 5 .

RemarKs
Our comment that the 'tens' digit of $2013^{2}-2013$ is the same as that of $13^{2}-13$ uses the fac that $2013^{2}-2013=(2000+13)^{2}-2013=\left(2000^{2}+2 \times 2000 \times 13+13^{2}\right)-(2000+13)=$ $4000000+52000-2000+13^{2}-13$. It is now clear that only the last two terms, that is, $13^{2}-13$ can have any effect on the 'tens' and 'units' digits of the answer.
A more sophisticated way to say this is to use the language and notation of modular arithmetic which you may already have met. Using this notation we write $a \equiv b(\bmod n)$ to mean tha $a$ and $b$ have the same remainder when divided by $n$. For example, $2013 \equiv 13(\bmod 100)$ and $156 \equiv 56(\bmod 100)$. Then we can say that $2013^{2}-2013 \equiv 13^{2}-13(\bmod 100)$ anc $13^{2}-13 \equiv 56(\bmod 100)$. It follows that $2013^{2}-2013 \equiv 56(\bmod 100)$. Thus $2013^{2}-2013$ has remainder 56 when divided by 100 . So its last two digits are 5 and 6 . In particular its 'tens digit is 5 and its 'units' digit is 6 .

FOR INVESTIGATION
3.1 Find the 'tens' digits of (a) $2014^{2}-2014$ and (b) $2013^{3}-2013^{2}$.
3.2 Find the 'tens' and 'units' digits of (a) $2011^{2011}$ and (b) $2013^{2013}$.