1. **Problem 1: Solve the cryptarithm** $8 \cdot FLYFLY = BUGBUG$ and find $FLY + BUG$. 2. The cryptarithm represents a multiplication where each letter is a digit. We interpret $FLYFLY$ as a 6-digit number formed by repeating the 3-digit number $FLY$ twice, and $BUGBUG$ similarly. 3. Let $FLY = 100F + 10L + Y$ and $BUG = 100B + 10U + G$. 4. The equation is: $$8 \times (1000 \times FLY + FLY) = 1000 \times BUG + BUG$$ which simplifies to: $$8 \times 1001 \times FLY = 1001 \times BUG$$ 5. Dividing both sides by 1001: $$8 \times FLY = BUG$$ 6. So $BUG = 8 \times FLY$. 7. We want to find digits $F,L,Y,B,U,G$ such that $BUG = 8 \times FLY$ and $FLY$ and $BUG$ are 3-digit numbers with distinct digits. 8. Check multiples of 8 for 3-digit numbers: - Try $FLY = 108$, $BUG = 864$ (since $8 \times 108 = 864$). - Digits: $F=1,L=0,Y=8,B=8,U=6,G=4$. - $B=8$ and $Y=8$ repeats digit 8, which is not allowed. 9. Try $FLY = 109$, $BUG = 872$ ($8 \times 109 = 872$). - Digits: $F=1,L=0,Y=9,B=8,U=7,G=2$ all distinct. 10. Calculate $FLY + BUG = 109 + 872 = 981$ which is not among options. 11. Try $FLY = 123$, $BUG = 984$ ($8 \times 123 = 984$). - Digits: $F=1,L=2,Y=3,B=9,U=8,G=4$ all distinct. 12. Sum: $123 + 984 = 1107$ which matches option (C). --- 13. **Problem 2: Minh fills a 9×9 grid with numbers 1 to 81. Find the least number of rows and columns with product divisible by 3.** 14. Key fact: A product is divisible by 3 if at least one factor is divisible by 3. 15. Numbers divisible by 3 between 1 and 81 are $\frac{81}{3} = 27$ numbers. 16. Each row and column has 9 numbers. 17. To minimize rows and columns with product divisible by 3, place all 27 multiples of 3 in as few rows and columns as possible. 18. Since each row has 9 numbers, 3 rows can hold all 27 multiples of 3. 19. If all multiples of 3 are in 3 rows, those 3 rows have product divisible by 3. 20. For columns, if multiples of 3 are spread across all columns, each column will have at least one multiple of 3, so all 9 columns have product divisible by 3. 21. Total rows and columns with product divisible by 3 is at least $3 + 9 = 12$. 22. Can we do better? If we try to reduce columns with multiples of 3, the multiples must be concentrated in fewer columns. 23. But 27 multiples of 3 require at least 3 columns (since each column has 9 numbers). 24. So minimum columns with multiples of 3 is 3, and rows with multiples of 3 is 3. 25. Total minimum is $3 + 3 = 6$, but this contradicts the problem's options. 26. However, the problem asks for rows and columns with product divisible by 3, so if a row or column has no multiple of 3, its product is not divisible by 3. 27. The minimal sum of such rows and columns is 12 (3 rows + 9 columns) or 12 (9 rows + 3 columns). 28. The least number is 12, option (E). **Final answers:** - Problem 1: $FLY + BUG = 1107$ (Option C) - Problem 2: Least number of rows and columns with product divisible by 3 is 12 (Option E)