## [高中]數與座標系

### [高中]數與座標系

1.設n是正整數且(n+1)∣(3n+14),求n﹖

2.設n是正整數，己知 n | 3n+10，求n?

3.設n是正整數，己知 (2n+3) | (6n-4)，求n?

(2n+3) | -1(6n-4)+3(2n+3)
(2n+3) | -13
2n+3>3 , 故2n+3=13 , 得n=5

arhauw

arhauw

(a).若 a | b 則 a | xb for all x in Z (Z為整數系)
(b).若 a | b 且 a | c 則 a | xb+yc for all x, y in Z

1.
(n+1) | (3n+14)
-> (n+1) | (3n+14)+k(n+1) for all k in Z
-> (n+1) | (3n+14)+(-3)(n+1) for k=-3
-> (n+1) | 11
因11為質數，故 n+1 = 1 or 11
又因n為正整數，故n+1 = 11
則 n=10

2.
n | 3n+10
-> n | 3n+10+kn for all k in Z
-> n | 3n+10+(-3)n for k=-3
-> n | 10
因n為正整數，則n為10的因數，即 n in {1, 2, 5, 10}

3.
(2n+3) | (6n-4)
-> (2n+3) | (6n-4)+k(2n+3) for all k in Z
-> (2n+3) | (6n-4)+(-3)(2n+3) for k=-3
-> (2n+3) | -13
-> (2n+3) | 13
因13為質數，故 2n+3 = 1 or 13
又因n為正整數，故2n+3 = 13
則 n = 5

Tzwan

for all k in Z

Z這樣嗎
----------
(n+1) | (3n+14)
-> (n+1) | (3n+14)+k(n+1) for all k in Z  (

-> (n+1) | (3n+14)+(-3)(n+1) for k=-3
(請問-3是怎麼出現的，-3是一個固定要帶入的數值嗎?因為K=-3這裡搞不太懂,而且這三題都有一樣的-3)

arhauw

for all 指 ∀
k in Z 指 k∈Z

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Tzwan