三個想不出來的數學題目?

1)

50!
= (5×10×15×20×25×30×35×40×45×50)
×(1×2×3×4)(6×7×8×9)(11×...×14)(16×...×19)(21×...×24)(26×...×29)(31×...34)(36×...×39)(41×...×44)(46×...×49)

= 5¹² (1×2×3×4)(6×7×8×9)×2
×(1×2×3×4)(6×7×8×9)(11×...×14)(16×...×19)(21×...×24)(26×...×29)(31×...34)(36×...×39)(41×...×44)(46×...×49)

= 10¹² (1×1×3×1)(3×7×1×9)×1
×(1×2×3×4)(3×7×1×9)(11×...×14)(16×...×19)(21×...×24)(26×...×29)(31×...34)(36×...×39)(41×...×44)(46×...×49)

50! 共有12個零結尾, 其最後"兩位"非零之數
= (1×1×3×1)(3×7×1×9)×1
×(1×2×3×4)(3×7×1×9)(11×...×14)(16×...×19)(21×...×24)(26×...×29)(31×...34)(36×...×39)(41×...×44)(46×...×49)
的最後兩位數字

= 3(189)
×(1×2×3×4)(189)(11×...×14)(16×...×19)(21×...×24)(26×...×29)(31×...34)(36×...×39)(41×...×44)(46×...×49)
的最後兩位數字

注意(5n+1)(5n+2)(5n+3)(5n+4)
≡ 5n(1×2×3 + 1×2×4 + 1×3×4 + 2×3×4) + 1×2×3×4 ≡ 250n + 24 ≡ 24(mod 25)
又 (5n+1)(5n+2)(5n+3)(5n+4) 是 4 的倍數, 故 (5n+1)(5n+2)(5n+3)(5n+4) ≡ 24(mod 100)

∴ 最後"兩位"非零之數 ≡ 3(189)² (24)^9 ≡ 3(21)(24)³ ≡ 3(21)(24) ≡ 12(mod 100)
最後"兩位"非零之數 = 12

2)

a³ + b³ + c³ = 0
a³ + b³ - (a+b)³ = 0
(a+b)(a²-ab+b²) - (a+b)³ = 0
(a+b)(a²-ab+b² - a²-2ab-b²) = 0
(a+b)ab = 0
- abc = 0
不妨設c = 0 , 則 a = - b,
a²º¹⁵ + b²º¹⁵ + c²º¹⁵ = (-b)²º¹⁵ + b²º¹⁵ + 0²º¹⁵= 0

3)

考慮 1/(2√(n+1)) < 1/(√n + √(n+1)) < 1/(2√n)
1/(2√(n+1)) < √(n+1) - √n < 1/(2√n)
1/√(n+1) < 2(√(n+1) - √n) < 1/√n

對 1/√(n+1) < 2(√(n+1) - √n) 取 n = 1 至 99 叠加得
1/√2 + 1/√3 + ... + 1/√100 < 2( (√2 - √1) + (√3 - √2) + ... + (√100 - √99) ) = 2(√100 - √1) = 18
1 + 1/√2 + 1/√3 + ... + 1/√100 < 19

對 2(√(n+1) - √n) < 1/√n 取 n = 1 至 99 叠加得
2( (√2 - √1) + (√3 - √2) + (√4 - √3) ... + (√100 - √99) ) = 2(√100 - √1) = 18 < 1/√1 + 1/√2 + 1/√3 + ... + 1/√99
18.1 < 1/√1 + 1/√2 + 1/√3 + ... + 1/√99 + 1/√100

綜合得 18.1 < 1/√1 + 1/√2 + 1/√3 + ... + 1/√99 + 1/√100 < 19
故其整數部分之值為 18.