數學一問.!!10分..thx.

First, you may substitute n into different numbers
For examples, I let n be 1, 2, 3............., 18, 19, 20
I find that when n = 3, 4, 10, 11, 17, 18
3² - 2 = 7(1)
4² - 2 = 7(2)
10² - 2 = 7( 14 )
11² - 2 = 7( 17 )
17² - 2 = 7( 41 )
18² - 2 = 7( 46 )
I divide them into 2 group ( n1 & n2 )
n1 = 3, 10, 17, 24........................
n2 = 4, 11, 18, 25........................
The pattern of n1 = 7(T) - 4 .........( T is the number of terms )
The pattern of n2 = 7(T) - 3..........( T is the number of terms )
In the first 2002 positive integers ,
(2002+4) ÷ 7 = 286............4
Because T must be an integer,
the highest value of T is 286
Because there are two groups,
286 x 2 = 572
572 positive integers are 「good 」



2007-04-13 15:37:03 補充：
下面果位朋友, 你求其列D數字出黎都估到唔會係16咁細~3, 4, 10, 11, 17, 18, 24, 25, 31, 32, 38, 39, 45, 46, 52, 53, 59, 60頭60個數字都已經有18個啦~

都幾難,,
∵Tn=(n^2-2)x7
T1=(1^2-2)x7=–7<<因為唔係整數所以不當
T2=(2^2-2)x7=14<<所以由2開始
T3=(3^2-2)x7=49
T4=(4^2-2)x7=98
.
.
.
.
.
T15=(15^2-2)x7=1561
T16=(16^2-2)x7=1792<<其實係到呢度就停不過寫多一個穩陣d
T17=(17^2-2)x7=2009<<多過左

∴16-1=15of the first 2002 positive integers are 「good 」

2007-04-13 18:27:15 補充：
牙!!~!謝上面果位提醒!~!我計錯架,,唔好跟我牙

572

(2002/7)*2=572