Question

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 1

Understanding

First, we are in base 10. Each digit value we write has powers of $10$ multiplied to it.

For example, $123=10^2+2\times10^1+3\times10^0$. 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!

Answer 2

Question:-

• What is the 'tens' digit of 2013^2-2013?

To Find:-

• Find tens place.

Solution:-

Here ,

Firstly , we need to find product of two integers. Then after we need to find the number.

Now ,

$\tt\implies \: { 13 }^{ 2 } - 13$

$\tt\implies \: 169 - 13$

$\tt\implies \: 156$

Hence ,

• The tens digit is 5.