What is the 'tens' digit of 2013^2-2013? (^ as in raised in power)(If

1.	What is the 'tens' digit of 2013^2-2013? (^ as in raised in power)(If you could explain too what the 'tens digit is because I'm not a native speaker and cannot understand it. I would really appreciate it! )
Answer» Understanding First, we are in BASE 10. Each digit value we write has powers of $10$ multiplied to it. For example, $123=10^2+2\times10^1+<klux>3</klux>\times10^0$ . If $10$ is multiplied, it is the TENS digit. Here tens digit value is 2. Sol. Now let's solve our problem. When we find a product of two integers, the first digit comes from below the tens digit. Similarly, the tens digit comes from below the hundreds digit. $13^2-13$ $=169-13$ $=156$ Hence the tens digit is $5$ . Question (I'm REALLY SORRY for leaving the wrong answer to your question.) The numbers p,q,r, and s satisfy the following equations: p+2q+3r+4s=k 4p=3q=2r=s What is the smallest value of 'k' for which p,q,r and are all positive integers? Sol. Here, $s$ is a multiple of 12 because of factors 4, 3, 2. The least value of $s=12$ , hence other values are $p=3, q=4, r=6$ . $\therefore k=3+8+18+48=77$ (A-20//B-24//C-25//D-77//E-154) Hope you understood!