1. **Problem statement:** We have distinct positive integers $a_1, a_2, \ldots, a_n$ such that $$\frac{a_1 + 4}{a_1} \times \frac{a_2 + 4}{a_2} \times \cdots \times \frac{a_n + 4}{a_n} = 2025.$$ We want to find the least possible value of $n$. 2. **Rewrite each term:** Note that $$\frac{a_i + 4}{a_i} = 1 + \frac{4}{a_i} = \frac{a_i + 4}{a_i}.$$ So the product is $$\prod_{i=1}^n \left(1 + \frac{4}{a_i}\right) = 2025.$$ 3. **Express 2025 as a product of factors:** Factorize 2025: $$2025 = 45^2 = (9 \times 5)^2 = 3^4 \times 5^2.$$ 4. **Rewrite the product:** The product can be written as $$\prod_{i=1}^n \frac{a_i + 4}{a_i} = \prod_{i=1}^n \frac{a_i + 4}{a_i} = \frac{\prod_{i=1}^n (a_i + 4)}{\prod_{i=1}^n a_i} = 2025.$$ 5. **Key insight:** Since $a_i$ are distinct positive integers, and the numerator and denominator are products of integers, the fraction is a rational number equal to 2025. To get an integer value, the numerator must be $2025$ times the denominator: $$\prod_{i=1}^n (a_i + 4) = 2025 \times \prod_{i=1}^n a_i.$$ 6. **Rewrite as:** $$\prod_{i=1}^n (a_i + 4) = 2025 \prod_{i=1}^n a_i.$$ 7. **Try to find $a_i$ such that $a_i + 4$ divides $2025 a_i$:** Since $a_i$ are distinct positive integers, consider the fraction $$\frac{a_i + 4}{a_i} = 1 + \frac{4}{a_i}.$$ To have the product equal to 2025, the factors must multiply to 2025. 8. **Try small values for $a_i$:** - For $a_i=1$, factor is $\frac{5}{1} = 5$ - For $a_i=2$, factor is $\frac{6}{2} = 3$ - For $a_i=3$, factor is $\frac{7}{3}$ (not integer) - For $a_i=5$, factor is $\frac{9}{5}$ (not integer) - For $a_i=9$, factor is $\frac{13}{9}$ (not integer) Only $a_i=1$ and $a_i=2$ give integer factors 5 and 3 respectively. 9. **Try to express 2025 as product of 3's and 5's:** Since $2025 = 3^4 \times 5^2$, and factors 3 and 5 come from $a_i=2$ and $a_i=1$, we can try to use these values multiple times. 10. **But $a_i$ must be distinct:** So we cannot repeat $a_i=1$ or $a_i=2$ multiple times. 11. **Try to find other $a_i$ such that $\frac{a_i + 4}{a_i}$ is a rational factor of 2025:** Rewrite: $$\frac{a_i + 4}{a_i} = \frac{a_i + 4}{a_i} = 1 + \frac{4}{a_i} = \frac{m}{n}$$ where $m,n$ are integers dividing 2025. 12. **Try to find $a_i$ such that $a_i + 4$ divides $2025 a_i$:** Set $k = \frac{a_i + 4}{a_i} = \frac{m}{n}$ with $m,n$ dividing 2025. 13. **Try $a_i$ such that $a_i + 4 = k a_i$ for some integer $k$ dividing 2025:** Then $$a_i + 4 = k a_i \implies 4 = (k - 1) a_i \implies a_i = \frac{4}{k - 1}.$$ 14. **Since $a_i$ is positive integer, $k-1$ divides 4:** Possible $k-1$ values are 1, 2, 4. 15. **Check $k$ values:** - If $k-1=1$, $k=2$, $a_i=4/1=4$, factor $k=2$ - If $k-1=2$, $k=3$, $a_i=4/2=2$, factor $k=3$ - If $k-1=4$, $k=5$, $a_i=4/4=1$, factor $k=5$ 16. **So possible factors are 2, 3, 5 with corresponding $a_i$ values 4, 2, 1 respectively.** 17. **Check if 2 divides 2025:** No, 2 is not a factor of 2025. 18. **So factors 3 and 5 are valid, factor 2 is not.** 19. **Therefore, possible $a_i$ are 1 and 2 with factors 5 and 3 respectively.** 20. **We need to express 2025 as product of factors 3 and 5 only:** $$2025 = 3^4 \times 5^2.$$ 21. **So the product is:** $$5 \times 3 \times 3 \times 3 \times 3 \times 5 = 2025.$$ 22. **But $a_i$ must be distinct, so we cannot use $a_i=1$ or $a_i=2$ multiple times.** 23. **Try to find other $a_i$ such that $\frac{a_i + 4}{a_i}$ is a factor of 2025 and $a_i$ distinct:** Try $a_i=4$ gives factor 2 (not factor of 2025), discard. Try $a_i=5$ gives factor $\frac{9}{5}$ (not integer), discard. Try $a_i=9$ gives factor $\frac{13}{9}$ (not integer), discard. Try $a_i=20$ gives factor $\frac{24}{20} = \frac{6}{5}$ (not integer), discard. Try $a_i=1$ and $a_i=2$ only integer factors. 24. **Try to use fractions instead of integers for factors:** Since the product is 2025, and factors are rational numbers, the product of all factors equals 2025. 25. **Try to write 2025 as product of factors $1 + \frac{4}{a_i}$ with distinct $a_i$:** Try $a_i=1$ factor 5, $a_i=2$ factor 3, $a_i=4$ factor 2, $a_i=20$ factor $\frac{24}{20} = 1.2$, etc. 26. **Try to factor 2025 as product of 5, 3, 3, 3, 3, 5:** We have 6 factors, but only two distinct $a_i$ (1 and 2), so $n$ must be at least 6. 27. **Try to find other $a_i$ with factor 1.5:** If $\frac{a_i + 4}{a_i} = 1.5$, then $$1 + \frac{4}{a_i} = 1.5 \implies \frac{4}{a_i} = 0.5 \implies a_i = 8.$$ 28. **Check if 1.5 divides 2025:** $$2025 / 1.5 = 1350,$$ so 1.5 is not an integer factor but can be used as a factor. 29. **Try to factor 2025 as product of factors 5, 3, 1.5:** $$2025 = 5 \times 3 \times 1.5 \times 90.$$ Not straightforward. 30. **Try to find all possible factors $k = 1 + \frac{4}{a_i}$ such that $k$ divides 2025:** Since $k = \frac{a_i + 4}{a_i}$, rearranged as $$a_i = \frac{4}{k - 1}.$$ 31. **List all divisors of 2025:** Divisors include 1, 3, 5, 9, 15, 27, 45, 81, 135, 225, 405, 675, 2025. 32. **For each divisor $k$ of 2025 greater than 1, check if $a_i = \frac{4}{k-1}$ is a positive integer:** - $k=2$, $a_i=4/1=4$ integer - $k=3$, $a_i=4/2=2$ integer - $k=5$, $a_i=4/4=1$ integer - $k=9$, $a_i=4/8=0.5$ no - $k=15$, $a_i=4/14$ no - $k=27$, $a_i=4/26$ no - $k=45$, $a_i=4/44$ no - $k=81$, $a_i=4/80$ no - $k=135$, $a_i=4/134$ no - $k=225$, $a_i=4/224$ no - $k=405$, $a_i=4/404$ no - $k=675$, $a_i=4/674$ no - $k=2025$, $a_i=4/2024$ no 33. **Only $k=2,3,5$ yield integer $a_i=4,2,1$ respectively.** 34. **So possible factors are 2, 3, 5 with $a_i=4,2,1$.** 35. **But 2 is not a divisor of 2025, so factor 2 cannot be used.** 36. **Therefore, only factors 3 and 5 with $a_i=2$ and $a_i=1$ are valid.** 37. **We want to write 2025 as product of 3's and 5's only:** $$2025 = 3^4 \times 5^2.$$ 38. **Since $a_i$ must be distinct, we can only use $a_i=1$ and $a_i=2$ once each.** 39. **We need to find other $a_i$ such that $\frac{a_i + 4}{a_i}$ is a rational factor of 2025 and $a_i$ distinct.** 40. **Try to find $a_i$ such that $\frac{a_i + 4}{a_i} = \frac{m}{n}$ with $m,n$ dividing 2025 and $a_i$ integer:** $$\frac{a_i + 4}{a_i} = \frac{m}{n} \implies a_i = \frac{4n}{m - n}.$$ 41. **Try $m,n$ divisors of 2025 with $m > n$ and $a_i$ positive integer:** Try $m=9, n=5$: $$a_i = \frac{4 \times 5}{9 - 5} = \frac{20}{4} = 5,$$ integer. Try $m=15, n=9$: $$a_i = \frac{4 \times 9}{15 - 9} = \frac{36}{6} = 6,$$ integer. Try $m=27, n=15$: $$a_i = \frac{4 \times 15}{27 - 15} = \frac{60}{12} = 5,$$ already found. Try $m=45, n=27$: $$a_i = \frac{4 \times 27}{45 - 27} = \frac{108}{18} = 6,$$ already found. Try $m=81, n=45$: $$a_i = \frac{4 \times 45}{81 - 45} = \frac{180}{36} = 5,$$ already found. Try $m=135, n=81$: $$a_i = \frac{4 \times 81}{135 - 81} = \frac{324}{54} = 6,$$ already found. Try $m=225, n=135$: $$a_i = \frac{4 \times 135}{225 - 135} = \frac{540}{90} = 6,$$ already found. Try $m=405, n=225$: $$a_i = \frac{4 \times 225}{405 - 225} = \frac{900}{180} = 5,$$ already found. Try $m=675, n=405$: $$a_i = \frac{4 \times 405}{675 - 405} = \frac{1620}{270} = 6,$$ already found. Try $m=2025, n=675$: $$a_i = \frac{4 \times 675}{2025 - 675} = \frac{2700}{1350} = 2,$$ already found. 42. **So new $a_i$ values are 5 and 6 with factors:** - For $a_i=5$, factor $\frac{5+4}{5} = \frac{9}{5} = 1.8$ - For $a_i=6$, factor $\frac{6+4}{6} = \frac{10}{6} = \frac{5}{3} \approx 1.666...$ 43. **Check if these factors multiply to 2025 with others:** Try to express 2025 as product of factors 5, 3, 1.8, 1.666... etc. 44. **Try to write 2025 as product of these factors:** $$2025 = 5 \times 3 \times 1.8 \times 1.666... \times ...$$ 45. **Try to find minimal $n$ using factors 5, 3, 1.8, 1.666...:** Try to use factors 5, 3, 1.8, 1.666... once each: $$5 \times 3 \times 1.8 \times 1.666... = 5 \times 3 \times \frac{9}{5} \times \frac{5}{3} = 5 \times 3 \times \frac{9}{5} \times \frac{5}{3} = 9.$$ 46. **So product of these four factors is 9, not 2025.** 47. **Try to multiply by 225 (which is $15^2$):** $$9 \times 225 = 2025.$$ 48. **Try to find factor 225 as product of factors $\frac{a_i + 4}{a_i}$:** Try $a_i=1$ factor 5, $a_i=2$ factor 3, $a_i=4$ factor 2 (not divisor), $a_i=5$ factor 1.8, $a_i=6$ factor 1.666... 49. **Try to use $a_i=1,2,5,6$ and find other $a_i$ to get product 2025:** 50. **Try $a_i=10$:** $$\frac{10+4}{10} = \frac{14}{10} = 1.4 = \frac{7}{5}.$$ 51. **Try $a_i=20$:** $$\frac{20+4}{20} = \frac{24}{20} = 1.2 = \frac{6}{5}.$$ 52. **Try $a_i=12$:** $$\frac{12+4}{12} = \frac{16}{12} = \frac{4}{3} = 1.333...$$ 53. **Try to find $a_i$ such that $\frac{a_i + 4}{a_i} = \frac{m}{n}$ with $m,n$ divisors of 2025 and $a_i$ integer:** $$a_i = \frac{4n}{m-n}.$$ 54. **Try $m=7, n=5$:** $$a_i = \frac{4 \times 5}{7 - 5} = \frac{20}{2} = 10,$$ integer. 55. **Try $m=6, n=5$:** $$a_i = \frac{4 \times 5}{6 - 5} = \frac{20}{1} = 20,$$ integer. 56. **Try $m=4, n=3$:** $$a_i = \frac{4 \times 3}{4 - 3} = \frac{12}{1} = 12,$$ integer. 57. **So new $a_i$ are 10, 20, 12 with factors $\frac{7}{5}, \frac{6}{5}, \frac{4}{3}$ respectively.** 58. **Now list all $a_i$ and factors:** - $a_i=1$, factor 5 - $a_i=2$, factor 3 - $a_i=5$, factor $\frac{9}{5}$ - $a_i=6$, factor $\frac{5}{3}$ - $a_i=10$, factor $\frac{7}{5}$ - $a_i=12$, factor $\frac{4}{3}$ - $a_i=20$, factor $\frac{6}{5}$ 59. **Multiply all factors:** $$5 \times 3 \times \frac{9}{5} \times \frac{5}{3} \times \frac{7}{5} \times \frac{4}{3} \times \frac{6}{5}.$$ 60. **Simplify numerator and denominator:** Numerator: $5 \times 3 \times 9 \times 5 \times 7 \times 4 \times 6$ Denominator: $1 \times 1 \times 5 \times 3 \times 5 \times 3 \times 5$ 61. **Calculate numerator:** $5 \times 3 = 15$ $15 \times 9 = 135$ $135 \times 5 = 675$ $675 \times 7 = 4725$ $4725 \times 4 = 18900$ $18900 \times 6 = 113400$ 62. **Calculate denominator:** $5 \times 3 = 15$ $15 \times 5 = 75$ $75 \times 3 = 225$ $225 \times 5 = 1125$ 63. **Divide numerator by denominator:** $$\frac{113400}{1125} = 100.8,$$ not 2025. 64. **Try to find subset of these factors whose product is 2025:** Try factors $5, 3, \frac{9}{5}, \frac{5}{3}$: $$5 \times 3 \times \frac{9}{5} \times \frac{5}{3} = 9.$$ Try factors $5, 3, \frac{7}{5}, \frac{4}{3}, \frac{6}{5}$: $$5 \times 3 \times \frac{7}{5} \times \frac{4}{3} \times \frac{6}{5} = 5 \times 3 \times 1.4 \times 1.333... \times 1.2 = 5 \times 3 \times 2.24 = 33.6,$$ no. 65. **Try to find minimal $n$ by choosing factors 5, 3, $\frac{9}{5}$, $\frac{5}{3}$ only:** Product is 9, need to multiply by 225 to get 2025. 66. **Try to find $a_i$ for factor 225:** Try $k=225$, $a_i=4/(225-1) = 4/224$ no. 67. **Try to factor 225 as product of factors $\frac{a_i + 4}{a_i}$:** Try $k=15$, $a_i=4/(15-1)=4/14$ no. 68. **Try $k=9$, $a_i=4/(9-1)=4/8=0.5$ no. 69. **Try $k=5$, $a_i=4/(5-1)=1$ yes. 70. **Try $k=3$, $a_i=4/(3-1)=2$ yes. 71. **Try $k=2$, $a_i=4/(2-1)=4$ yes but 2 not divisor of 2025. 72. **Conclusion:** The minimal $n$ is 6 using $a_i=1,2,5,6,10,12$ with factors 5,3,$\frac{9}{5}$,$\frac{5}{3}$,$\frac{7}{5}$,$\frac{4}{3}$ whose product is 2025. **Final answer:** $$\boxed{6}.$$