已知y=ax+b(a為整數)通過(98,19),
它的x軸交點為(p,0),y軸交點為(0,q)
p是質數,q是正整數
滿足條件的一次函數有幾多個?
一次函數的問題
2007-05-25 1:39 am
回答 (2)
2007-05-25 3:05 am
✔ 最佳答案
The equation can be written like this:19 = 98a + b
0 = pa + b
q = b
須符合以上三個條件
q = -pa
19 = 98a - pa
19 = a(98-p)
a = 19 , 98-p = 1 --> p = 97 (rejected q = -ve)
or 98-p = 19 a = 1 ---> p = 79 (rejected q = -ve)
a = -19 , 98-p = -1 ---> p= 99 (rejected)
a = -1 , 98-p = -19 --> p = 117 (rejected)
Therefore 滿足條件的一次函數 有 0 個
參考: me
2007-05-25 10:22 pm
19 = 98a + b
b = 19 - 98a
x - in : -b/a
y - in : b
-b/a = p and b = q
so,a = 1 or a = -1
-b = p or b = p
-(19 - 98(1)) = p or 19 - 98(-1) = p
p = 79 or p = 117(rejected since 117 is divisible by 3)
q = 79
there is only one possible linear equation
b = 19 - 98a
x - in : -b/a
y - in : b
-b/a = p and b = q
so,a = 1 or a = -1
-b = p or b = p
-(19 - 98(1)) = p or 19 - 98(-1) = p
p = 79 or p = 117(rejected since 117 is divisible by 3)
q = 79
there is only one possible linear equation
參考: eason mensa
收錄日期: 2021-04-12 19:43:06
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