1. **State the problem:** We need to find a four-digit number $ABCD$ where each digit is a perfect square (0, 1, 4, 9). 2. **Given conditions:** - The units digit $D$ is the sum of the thousands $A$, hundreds $B$, and tens $C$ digits: $$D = A + B + C$$ - The thousands digit $A$ is the product of the tens $C$ and hundreds $B$ digits: $$A = B \times C$$ - The sum of the thousands $A$ and tens $C$ digits is one less than the units digit $D$: $$A + C = D - 1$$ 3. **Analyze the conditions:** From the first and third equations: $$D = A + B + C$$ $$A + C = D - 1$$ Substitute $D$ from the first into the second: $$A + C = (A + B + C) - 1$$ Simplify: $$A + C = A + B + C - 1$$ Subtract $A + C$ from both sides: $$0 = B - 1$$ So: $$B = 1$$ 4. **Use $B=1$ in the second condition:** $$A = B \times C = 1 \times C = C$$ So $A = C$. 5. **Use $A = C$ and $B=1$ in the first condition:** $$D = A + B + C = A + 1 + A = 2A + 1$$ 6. **Use $A + C = D - 1$ with $A = C$:** $$A + A = D - 1$$ $$2A = D - 1$$ Substitute $D = 2A + 1$: $$2A = (2A + 1) - 1$$ $$2A = 2A$$ This is consistent. 7. **Digits must be perfect squares:** Possible digits: 0, 1, 4, 9. Since $A = C$ and $B=1$, try $A = C$ as 0, 1, 4, or 9. - If $A = 0$, then $D = 2(0) + 1 = 1$. Number: $0 1 0 1$ but $A=0$ means the number is not four-digit. - If $A = 1$, then $D = 2(1) + 1 = 3$ (3 is not a perfect square). - If $A = 4$, then $D = 2(4) + 1 = 9$ (9 is a perfect square). Number: $4 1 4 9$ - If $A = 9$, then $D = 2(9) + 1 = 19$ (not a digit). 8. **Check the number 4149:** - Digits are perfect squares: 4, 1, 4, 9 ✓ - $D = A + B + C = 4 + 1 + 4 = 9$ ✓ - $A = B \times C = 1 \times 4 = 4$ ✓ - $A + C = 4 + 4 = 8$ and $D - 1 = 9 - 1 = 8$ ✓ **Answer:** The number is **4149**.