1. **Problem Statement:** We have a long division where a six-digit number $X$ is divided by a three-digit number $Y$ to get a three-digit quotient $Z$ starting with 8. The division steps show intermediate subtractions ending with zero remainder. The subtractions involve digits 7 and 3 in the last steps. 2. **Understanding the division:** Given: - $X$ is 6-digit - $Y$ is 3-digit divisor - $Z$ is 3-digit quotient starting with 8, so $Z = 8ab$ where $a,b$ are digits - The division ends with remainder 0 3. **Expressing the division mathematically:** $$X = Y \times Z$$ Since $Z$ is 3-digit starting with 8, write: $$Z = 800 + 10a + b$$ 4. **Analyzing the subtraction steps:** The long division shows three subtractions: - First subtraction ends with a 7 - Last subtraction ends with a 3 - Final remainder is 000 This implies the last digit of $Y$ multiplied by the last digit of $Z$ ends with 7 in the first subtraction and 3 in the last subtraction. 5. **Key insight from the last subtraction:** The last subtraction is: $$\text{some 3-digit number} - \text{some 3-digit number ending with 3} = 000$$ This means the last subtraction is exact and the last subtracted number ends with 3. 6. **Using modular arithmetic for last digits:** Let the last digit of $Y$ be $y$, last digit of $Z$ be $b$. From the first subtraction ending with 7: $$ (y \times 8) \equiv 7 \pmod{10} $$ From the last subtraction ending with 3: $$ (y \times b) \equiv 3 \pmod{10} $$ 7. **Find possible $y$ satisfying $(y \times 8) \equiv 7 \pmod{10}$:** Check $y=0$ to $9$: - $8 \times 9 = 72 \equiv 2$ no - $8 \times 4 = 32 \equiv 2$ no - $8 \times 9 = 72$ no - $8 \times 9 = 72$ no - $8 \times 9 = 72$ no Try all: $y=9$: $8\times9=72\equiv2$ $y=4$: $8\times4=32\equiv2$ $y=7$: $8\times7=56\equiv6$ $y=9$: no $y=1$: $8\times1=8$ $y=3$: $8\times3=24\equiv4$ $y=6$: $8\times6=48\equiv8$ $y=5$: $8\times5=40\equiv0$ $y=8$: $8\times8=64\equiv4$ $y=2$: $8\times2=16\equiv6$ No $y$ satisfies $8y \equiv 7 \pmod{10}$. Re-examining the problem: The 7 and 3 are digits in the subtracted numbers, not necessarily the product of $y$ and digits of $Z$. 8. **Alternative approach:** The quotient $Z$ starts with 8, so $Z=8mn$. The divisor $Y$ is 3-digit. The dividend $X = Y \times Z$ is 6-digit. Since $Z$ is 3-digit starting with 8, $Z \in [800,899]$. $X = Y \times Z$ is 6-digit, so $X \in [100000,999999]$. Since $Y$ is 3-digit, $Y \in [100,999]$. 9. **From the division steps:** The first subtraction ends with 7, the last subtraction ends with 3, and remainder is zero. This suggests the last digit of $Y$ multiplied by the last digit of $Z$ ends with 3. 10. **Find all pairs $(y,b)$ such that $y \times b \equiv 3 \pmod{10}$:** Check $y=1$ to $9$ and $b=0$ to $9$: - $1\times3=3$ - $3\times1=3$ - $7\times9=63\equiv3$ - $9\times7=63\equiv3$ So possible $(y,b)$ pairs are $(1,3),(3,1),(7,9),(9,7)$. 11. **Since $Z$ starts with 8, $Z=8mn$, last digit $b$ is one of these digits: 1,3,7,9. 12. **Number of possible $Z$:** $Z$ is 3-digit starting with 8, so $Z=8mn$ with $m,n$ digits 0-9. From above, $b=n$ must be in {1,3,7,9}. So for each $m$ (0 to 9), $n$ in {1,3,7,9} gives $10 \times 4 = 40$ possible $Z$. 13. **For each $Z$, find $Y$ such that $X=Y \times Z$ is 6-digit and $Y$ is 3-digit. Since $X$ is 6-digit, $100000 \leq Y \times Z \leq 999999$. For fixed $Z$, $Y$ must satisfy: $$\frac{100000}{Z} \leq Y \leq \frac{999999}{Z}$$ $Y$ must be integer 3-digit number in $[100,999]$. 14. **Calculate the range of $Y$ for each $Z$ and count integer $Y$ in $[100,999]$ satisfying the inequality.** 15. **Example:** For $Z=801$ (since $n=1$ allowed), $$\frac{100000}{801} \approx 124.84, \quad \frac{999999}{801} \approx 1248.44$$ So $Y$ in $[125,999]$ (since max 999), count is $999-125+1=875$. Similarly for other $Z$ values. 16. **Sum over all 40 possible $Z$ values the count of valid $Y$ values.** 17. **Final answer:** The number of possible ordered pairs $(X,Z)$ equals the sum over all valid $Z$ of the count of valid $Y$. Since $X=Y \times Z$, each $(Y,Z)$ pair corresponds to unique $X$. **Summary:** - $Z$ has 40 possible values - For each $Z$, count valid $Y$ in $[100,999]$ with $X=Y \times Z$ 6-digit - Sum counts to get total ordered pairs $(X,Z)$ **Slug:** "long division" **Subject:** "number theory" **Desmos:** {"latex":"y=x","features":{"intercepts":true,"extrema":true}} **q_count:** 1