1. **State the problem:** We need to find a three-digit year where the digits satisfy the following conditions: - The tens digit is 1 less than the units digit. - If the tens and units digits were sides of a rectangle, its area would be 6 square units. - The hundreds digit is 300% (3 times) of the units digit. - The sum of all digits equals the sum of the first five counting numbers. 2. **Write down the variables and equations:** Let the hundreds digit be $h$, the tens digit be $t$, and the units digit be $u$. From the problem: - $t = u - 1$ - Area of rectangle formed by $t$ and $u$ is 6, so $t \times u = 6$ - $h = 3u$ - Sum of digits $h + t + u = 1 + 2 + 3 + 4 + 5 = 15$ 3. **Use the first two equations to find possible $u$ and $t$:** Substitute $t = u - 1$ into $t \times u = 6$: $$ (u - 1) \times u = 6 $$ $$ u^2 - u = 6 $$ $$ u^2 - u - 6 = 0 $$ 4. **Solve the quadratic equation:** $$ u = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2} $$ Possible values: - $u = \frac{1 + 5}{2} = 3$ - $u = \frac{1 - 5}{2} = -2$ (not a digit) So, $u = 3$. 5. **Find $t$ and $h$:** $$ t = u - 1 = 3 - 1 = 2 $$ $$ h = 3u = 3 \times 3 = 9 $$ 6. **Check the sum of digits:** $$ h + t + u = 9 + 2 + 3 = 14 $$ But the sum should be 15, so this does not satisfy the sum condition. 7. **Re-examine the sum condition:** Sum of first five counting numbers is $1 + 2 + 3 + 4 + 5 = 15$. Try the other possible $u$ values that satisfy digit constraints (0 to 9) and area condition. 8. **Try $u=6$:** If $u=6$, then $t = u - 1 = 5$. Check area: $$ t \times u = 5 \times 6 = 30 \neq 6 $$ No. 9. **Try $u=1$:** Then $t = 0$, area: $$ 0 \times 1 = 0 \neq 6 $$ No. 10. **Try $u= -2$ (discarded), $u=3$ (already tried), try $u= -1$ (discarded).** 11. **Try $u= 2$:** Then $t = 1$, area: $$ 1 \times 2 = 2 \neq 6 $$ No. 12. **Try $u= 4$:** Then $t = 3$, area: $$ 3 \times 4 = 12 \neq 6 $$ No. 13. **Try $u= 6$ (already tried), $u= 7$:** Then $t=6$, area: $$ 6 \times 7 = 42 \neq 6 $$ No. 14. **Try $u= 1$ to $9$ systematically:** Only $u=3$ and $t=2$ satisfy area 6. 15. **Check sum condition with $h=3u=9$, $t=2$, $u=3$:** Sum is 14, not 15. 16. **Check if hundreds digit can be 300% of units digit meaning $h=3u$ but $h$ must be a digit (0-9).** If $u=4$, $h=12$ invalid. If $u=3$, $h=9$ valid. 17. **Try sum of digits = 14 instead of 15:** Sum of first five counting numbers is 15, so maybe a typo or check if sum is 14. 18. **Conclusion:** The only digits satisfying all but sum condition are $h=9$, $t=2$, $u=3$. **Final answer:** The year is **923**. --- **Note:** The problem states sum of digits equals sum of first five counting numbers (15), but no digits satisfy all conditions with sum 15. The closest valid solution is 923 with sum 14.