A four-digit hexadecimal integer is written on a napkin such that the units digit is illegible. The first three digits are $7$, $D$, and $3$. If the integer is a multiple of $25_{10}$, what is the units digit?

Answer 1

To solve the problem, we need to determine the units digit \( x \) of a four-digit hexadecimal integer \( 7D3x \) such that this number is a multiple of \( 25_{10} \).

First, convert the known part of the hexadecimal number \( 7D3 \) to its decimal form. In hexadecimal, \( 7D3 \) represents:

\[ 7 \times 16^2 + D \times 16 + 3 \]

Since \( D \) corresponds to \( 13_{10} \):

\[
7 \times 256 + 13 \times 16 + 3 = 7 \times 256 + 13 \times 16 + 3
\]

Calculate each term:

\[
7 \times 256 = 1792
\]
\[
13 \times 16 = 208
\]

Adding these together:

\[
1792 + 208 + 3 = 2003
\]

Thus, the known part of the hexadecimal number \( 7D3 \) converts to \( 2003_{10} \).

Next, consider the full hexadecimal number \( 7D3x \), where \( x \) is a digit from 0 to 15. For the number \( 7D3x \) to be a multiple of \( 25_{10} \), the decimal equivalent \( 2003 + x \) must be a multiple of \( 25_{10} \).

Examining this, \( 2003 \div 25 \) provides:

\[
2003 \div 25 = 80, \text{ remainder } 3
\]

This means:

\[
2003 \equiv 3 \pmod{25}
\]

For \( 2003 + x \) to be a multiple of 25, \( x \) must satisfy:

\[
2003 + x \equiv 0 \pmod{25}
\]

Given \( 2003 \equiv 3 \pmod{25} \):

\[
3 + x \equiv 0 \pmod{25}
\]

Solving for \( x \):

\[
x \equiv -3 \pmod{25}
\]
\[
x \equiv 22 \pmod{25}
\]

Since \( x \) must be a single hexadecimal digit, it can only be 0 to 15. Since 22 is not a valid hexadecimal digit, we need to recheck our assumptions and notes.

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