1. **State the problem:** We need to fill in the missing digits in the multiplication puzzle: $$\boxed{\square \square 5} \times 4 \square = \square \square \square \square$$ with partial products: $$\square 1 5$$ $$\square 6 \square 0$$ $$4 7 \square \square$$ 2. **Understand the multiplication layout:** The multiplication is of a three-digit number ending with 5 by a two-digit number starting with 4. 3. **Set variables:** Let the three-digit number be $100a + 10b + 5$ where $a,b$ are digits. Let the two-digit number be $40 + c$ where $c$ is a digit. 4. **Partial products:** Multiplying by $c$ (units digit of multiplier) gives the first partial product ending with 5: $$(100a + 10b + 5) \times c = \square 1 5$$ This means the product ends with 5, so $c$ must be 1 or 5 (since only 1 or 5 multiplied by 5 ends with 5). 5. **Try $c=1$:** $$(100a + 10b + 5) \times 1 = 100a + 10b + 5 = \square 1 5$$ The last two digits are 1 and 5, so $10b + 5$ ends with 15, so $b=1$. So the first partial product is $a15$. 6. **Second partial product:** Multiplying by 4 (tens digit) gives: $$(100a + 10b + 5) \times 4 = \square 6 \square 0$$ This product ends with 0, so the units digit of $(100a + 10b + 5) \times 4$ is 0. Since the original number ends with 5, $5 \times 4 = 20$, units digit 0, consistent. 7. **Third partial product:** The sum of partial products is: $$\square \square \square \square = 4 7 \square \square$$ The total product starts with 4 and 7. 8. **Try $c=5$:** $$(100a + 10b + 5) \times 5 = \square 1 5$$ Since $5 \times 5 = 25$, units digit 5, tens digit 2, so the last two digits are 25, not 15. So $c=5$ is invalid. 9. **From step 5, $c=1$, $b=1$.** So the three-digit number is $100a + 10 \times 1 + 5 = 100a + 15$. 10. **Calculate second partial product:** $$(100a + 15) \times 4 = 400a + 60$$ This must be $\square 6 \square 0$. So the tens digit is 6, units digit 0, consistent. 11. **Calculate total product:** $$(100a + 15) \times 41 = (100a + 15) \times (40 + 1) = (100a + 15) \times 40 + (100a + 15) \times 1 = 4000a + 600 + 100a + 15 = 4100a + 615$$ This total product is $4 7 \square \square$. 12. **Try values of $a$ from 1 to 9:** - For $a=1$: total = $4100 + 615 = 4715$ which matches $4 7 1 5$. - Check partial products: First partial product: $(100 \times 1 + 15) \times 1 = 115$ matches $\square 1 5$. Second partial product: $(100 \times 1 + 15) \times 4 = 460$ matches $\square 6 \square 0$. Total product: $4715$ matches $4 7 \square \square$. 13. **Final answer:** The missing digits are: Three-digit number: 115 Two-digit number: 41 Product: 4715 **Verification:** $$115 \times 41 = 4715$$ All partial products and digits match the puzzle. **Answer:** The missing digits are 1 and 1 in the first number and 1 in the second number, making the multiplication $115 \times 41 = 4715$.