Question:
For the binary operation \( \times_{10} \) on the set \( S = \{1,3,7,9\} \), find the inverse of 3.
Concept:
The inverse of an element \( a \) under multiplication modulo 10 is an element \( b \in S \) such that:
\[ a \times_{10} b \equiv 1 \pmod{10} \]
Solution:
Step 1: We need to find \( b \in \{1,3,7,9\} \) such that
\[ 3 \times b \equiv 1 \pmod{10} \]
Step 2: Check elements of the set:
- \( 3 \times 1 = 3 \equiv 3 \pmod{10} \)
- \( 3 \times 3 = 9 \equiv 9 \pmod{10} \)
- \( 3 \times 7 = 21 \equiv 1 \pmod{10} \) ✅
- \( 3 \times 9 = 27 \equiv 7 \pmod{10} \)
Step 3: Since
\[ 3 \times 7 \equiv 1 \pmod{10} \]
Therefore, 7 is the inverse of 3.
Final Answer:
\[ \boxed{7} \]