1. **State the problem:** We need to find a two-digit number divisible by 4, whose digits add up to 10, and when reversed, the new number is smaller than the original by 18. 2. **Define variables:** Let the two-digit number be $10x + y$, where $x$ is the tens digit and $y$ is the units digit. 3. **Write down the conditions:** - Divisible by 4: $10x + y$ is divisible by 4. - Digits add up to 10: $x + y = 10$. - Reversed number is smaller by 18: $(10x + y) - (10y + x) = 18$. 4. **Simplify the third condition:** $$ (10x + y) - (10y + x) = 18 \implies 10x + y - 10y - x = 18 \implies 9x - 9y = 18 \implies x - y = 2 $$ 5. **Solve the system of equations:** From $x + y = 10$ and $x - y = 2$, add both: $$ (x + y) + (x - y) = 10 + 2 \implies 2x = 12 \implies x = 6 $$ Substitute $x=6$ into $x + y = 10$: $$ 6 + y = 10 \implies y = 4 $$ 6. **Check divisibility by 4:** Number is $10x + y = 10 \times 6 + 4 = 64$. Since $64 \div 4 = 16$, it is divisible by 4. 7. **Verify the reversed number condition:** Reversed number is $10y + x = 10 \times 4 + 6 = 46$. Difference: $64 - 46 = 18$, which matches the condition. **Final answer:** The number is **64**.