整數除法的問題

Question

n,a都是正整數，a≧4，n除以a所得的餘數是x，n除以(a+1)所得的餘數是(x-3)，求n除以[a(a+1)]所得的餘數。

Answer 1

設 n = ap + x = (a+1)q + x - 3, 則 n - x = ap = (a+1)q - 3.
令 n - x = a(a+1)Q + R , 則 R = (a+1)K - 3 = ak , 有 (a+1)K = ak + 3 , 得 K = k = 3.
故 n - x = a(a+1)Q + 3a , 即 n = a(a+1)Q + 3a+x , 又 a≧4 且a > x , 故 a(a+1) ≧ 5a > 4a > 3a+x ,
所以 n 除以 [a(a+1)] 所得的餘數是 3a+x.