幾題高中數學問題(因式和整數系單元)?

1)
2 × 4 × 6 × ...× 200 = 10k × P
100! × 2¹ºº = 10k × P
k = 100!/10 × 2¹ºº / P

因[100/2] + [100/4] + [100/8] + [100/16] + [100/32] + [100/64] = 50 + 25 + 12 + 6 + 3 + 1 = 97
[100/5] + [100/25] = 20 + 4 = 24
故
k = 2^97 × 5^24 × ... /10 × 2¹ºº / P
k = 2^196 × 5²³ × ... / P , 且10不整除P , 則
k = 2^196 × ... / P 或 k = 5²³ × ... / P (兩者可同時成立)。

2)
因 9x + 5y 與10x + ky 都是 11 的倍數，故 9kx + 5ky - (50x + 5ky) = (9k - 50)x 是 11 的倍數。
x 非 11 的倍數得 9k - 50 是 11 的倍數, k 最小= 8 , 有 9(8) + 5(1) = 77 , 10(8) + 8(1) = 88 皆是 11 的倍數。

3)
3a + 2b = g + l
3a + 2b = g + ab/g
3a/g + 2b/g = 1 + a/g b/g
令 a/g = p ≤ b/g = q , 則 (p,q) = 1 ,
3p + 2q - pq = 1
(p - 2)(- q + 3) + 6 = 1
(p - 2)(q - 3) = 5

p - 2 = - 5 , q - 3 = - 1 ⇒ p = - 3 < 0 (捨去)
p - 2 = 1 , q - 3 = 5 ⇒ p = 3 , q = 8
則 a = 3g , b = 8g , 得 20 ≤ 3g ≤ 8g ≤ 60 ⇒ g = 7 ,
故 a = 21 , b = 56。

4)
ab = 15!
= 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 × 12 × 13 × 14 × 15
= 2 × 3 × 2² × 5 × (2×3) × 7 × 2³ × 3² × (2×5) × 11 × (2²×3) × 13 × (2×7) × 3×5
= 2¹¹ × 3^6 × 5³ × 7² × 11 × 13
(a,b)之最大值 = 2^5 × 3³ × 5 × 7 = 30240。