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, hundreds, and tens digits: $$D = A + B + C$$ - The thousands digit $A$ is the product of the tens and hundreds digits: $$A = C \times B$$ - The sum of the thousands and tens digits is one less than the units digit: $$A + C = D - 1$$ 3. **Analyze the digits:** Since each digit is a perfect square, possible digits are $0,1,4,9$. 4. **Use the equations:** 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, $$B = 1$$ 5. **Use $B=1$ in $A = C \times B$:** $$A = C \times 1 = C$$ 6. **Rewrite $D = A + B + C$ with $A=C$ and $B=1$:** $$D = C + 1 + C = 2C + 1$$ 7. **Recall $A + C = D - 1$ and substitute $A=C$ and $D=2C+1$:** $$C + C = (2C + 1) - 1$$ $$2C = 2C$$ This is always true, so no new info. 8. **Check possible values for $C$ (perfect squares):** Possible $C$ values: 0, 1, 4, 9 - If $C=0$: - $A = 0$ - $D = 2(0) + 1 = 1$ - Number: $A B C D = 0 1 0 1$ (not a four-digit number, leading zero invalid) - If $C=1$: - $A = 1$ - $D = 2(1) + 1 = 3$ (3 is not a perfect square, invalid) - If $C=4$: - $A = 4$ - $D = 2(4) + 1 = 9$ (9 is a perfect square) - Number: $4 1 4 9$ - If $C=9$: - $A = 9$ - $D = 2(9) + 1 = 19$ (not a digit, invalid) 9. **Verify the number 4149:** - Digits are perfect squares: 4,1,4,9 ✓ - $D = A + B + C = 4 + 1 + 4 = 9$ ✓ - $A = C \times B = 4 \times 1 = 4$ ✓ - $A + C = D - 1 \Rightarrow 4 + 4 = 9 - 1 \Rightarrow 8 = 8$ ✓ **Final answer:** The number M.O.B is thinking of is **4149**.