1.設P為質數且(P^3+3PP-4P+40)/(P-1)也是質數,求P值
2.自然數N被5除餘3,4N被6除餘2,2N被7除餘5,最接近一萬之自然數為?
3.n為整數,若n^4-38nn+169為質數,求n之值?
4.設方程式XX-(a-24)x+(a-1)=0之二根均為整數,且a為自然數,試求a之值
因數與倍數與根與係數
版主: thepiano
Re: 因數與倍數與根與係數
第 1 題
(p^3 + 3p^2 - 4p + 40) / (p - 1)
= (p^3 + 3p^2 - 4p) / (p - 1) + 40 / (p - 1)
= p(p + 4) + 40 / (p - 1)
故 p - 1 是 40 的正因數且為質數
p = 2,3,5,11,41
一一代回 p(p + 4) + 40 / (p - 1) 檢驗可知 p = 3
第 2 題
n ≡ 3 (mod 5)
4n ≡ 2 (mod 6)
4n = 8,20,32,44,......
n = 2,5,8,11,......
n ≡ 2 (mod 3)
2n ≡ 5 (mod 7)
2n = 12,26,40,54,......
n = 6,13,20,27,......
n ≡ 6 (mod 7)
n ≡ 2 (mod 3) 和 n ≡ 6 (mod 7)
可知 n = 21 之倍數 - 1
接近 10000 的 (21 之倍數 - 1)是 10058,10037,10016,9995,9953,......
再由 n ≡ 3 (mod 5)
可知所求為 9953
第 3 題
http://forum.nta.org.tw/examservice/sho ... php?t=1575
第 4 題
題目有沒有打錯?
(p^3 + 3p^2 - 4p + 40) / (p - 1)
= (p^3 + 3p^2 - 4p) / (p - 1) + 40 / (p - 1)
= p(p + 4) + 40 / (p - 1)
故 p - 1 是 40 的正因數且為質數
p = 2,3,5,11,41
一一代回 p(p + 4) + 40 / (p - 1) 檢驗可知 p = 3
第 2 題
n ≡ 3 (mod 5)
4n ≡ 2 (mod 6)
4n = 8,20,32,44,......
n = 2,5,8,11,......
n ≡ 2 (mod 3)
2n ≡ 5 (mod 7)
2n = 12,26,40,54,......
n = 6,13,20,27,......
n ≡ 6 (mod 7)
n ≡ 2 (mod 3) 和 n ≡ 6 (mod 7)
可知 n = 21 之倍數 - 1
接近 10000 的 (21 之倍數 - 1)是 10058,10037,10016,9995,9953,......
再由 n ≡ 3 (mod 5)
可知所求為 9953
第 3 題
http://forum.nta.org.tw/examservice/sho ... php?t=1575
第 4 題
題目有沒有打錯?
Re: 因數與倍數與根與係數
第 4 題
令兩根為 p 和 q
p + q = a - 24
pq = a - 1
(1 - p)(1 - q) = 1 - (p + q) + pq = 24
1 - p = 24,1 - q = 1
1 - p = 12,1 - q = 2
1 - p = 8,1 - q = 3
1 - p = 6,1 - q = 4
1 - p = -24,1 - q = -1
1 - p = -12,1 - q = -2
1 - p = -8,1 - q = -3
1 - p = -6,1 - q = -4
a = pq + 1 = 1,12,15,16,36,37,40,51
令兩根為 p 和 q
p + q = a - 24
pq = a - 1
(1 - p)(1 - q) = 1 - (p + q) + pq = 24
1 - p = 24,1 - q = 1
1 - p = 12,1 - q = 2
1 - p = 8,1 - q = 3
1 - p = 6,1 - q = 4
1 - p = -24,1 - q = -1
1 - p = -12,1 - q = -2
1 - p = -8,1 - q = -3
1 - p = -6,1 - q = -4
a = pq + 1 = 1,12,15,16,36,37,40,51