Subtract the following hexadecimal numbers: EBA16 - 2E16

We need to calculate $A_{16} - E_{16}$.
In decimal, this is $10 - 14$. Since $10 < 14$, we need to borrow from the next column.
Borrow 1 from B in the $16^1$ place. B ($11_{10}$) becomes A ($10_{10}$).
The borrowed 1 is equivalent to $16_{10}$ in the current column. So, A ($10_{10}$) becomes $10 + 16 = 26_{10}$.
Now, subtract: $26_{10} - E_{16} = 26_{10} - 14_{10} = 12_{10}$.
In hexadecimal, $12_{10}$ is C.
So, the rightmost digit is C.

The B in $EBA_{16}$ became A after borrowing.
We need to calculate $A_{16} - 2_{16}$.
In decimal, this is $10 - 2 = 8_{10}$.
In hexadecimal, $8_{10}$ is 8.
So, the middle digit is 8.

We need to calculate $E_{16} - 0_{16}$ (since there's no digit in the $16^2$ place for $2E_{16}$).
In decimal, this is $14 - 0 = 14_{10}$.
In hexadecimal, $14_{10}$ is E.
So, the leftmost digit is E.

Combining the results:

Problem (d):

We'll subtract column by column, from right to left.

We need to calculate $A_{16} - A_{16}$.
In decimal, this is $10 - 10 = 0_{10}$.
So, the rightmost digit is 0.

We need to calculate $B_{16} - B_{16}$.
In decimal, this is $11 - 11 = 0_{10}$.
So, the second digit from the right is 0.

We need to calculate $A_{16} - E_{16}$.
In decimal, this is $10 - 14$. Since $10 < 14$, we need to borrow from the next column.
Borrow 1 from B in the $16^3$ place. B ($11_{10}$) becomes A ($10_{10}$).
The borrowed 1 is equivalent to $16_{10}$ in the current column. So, A ($10_{10}$) becomes $10 + 16 = 26_{10}$.
Now, subtract: $26_{10} - E_{16} = 26_{10} - 14_{10} = 12_{10}$.
In hexadecimal, $12_{10}$ is C.
So, the third digit from the right is C.

The B in $BABA_{16}$ became A after borrowing.
We need to calculate $A_{16} - 0_{16}$ (since there's no digit in the $16^3$ place for $EBA_{16}$).
In decimal, this is $10 - 0 = 10_{10}$.
In hexadecimal, $10_{10}$ is A.
So, the leftmost digit is A.

Combining the results: