cos1 +cos3+cos5+....+cos359
用cos(180-X) = - cos X 拆兩件抵銷
>cos1...+cos179+cos(180+1)..+cos+(180+179)
呢到就+到179
(sin2)^2+(sin4)^2+...+(sin90)^2
用sin^2+cos^2=1
同sin^2(90-X) = cos^2 X解
拆開中間數字就+到44
應該係小學 單偶數既問題? 應該點搵 係咪最大數字除2 退1格?
冇讀好多年書 好多基本野都唔識 自修緊 可否幫幫手
[DSE 2021] 大家準備好瘋狂做 數學 未呢？ [5]
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想問象限問題 拆兩件抵銷 點睇中間既數字+到去邊?
cos1 +cos3+cos5+....+cos359
用cos(180-X) = - cos X 拆兩件抵銷
>cos1...+cos179+cos(180+1)..+cos+(180+179)
呢到就+到179
(sin2)^2+(sin4)^2+...+(sin90)^2
用sin^2+cos^2=1
同sin^2(90-X) = cos^2 X解
拆開中間數字就+到44
應該係小學 單偶數既問題? 應該點搵 係咪最大數字除2 退1格?
冇讀好多年書 好多基本野都唔識 自修緊 可否幫幫手
cos1 +cos3+cos5+....+cos359
用cos(180-X) = - cos X 拆兩件抵銷
>cos1...+cos179+cos(180+1)..+cos+(180+179)
呢到就+到179
(sin2)^2+(sin4)^2+...+(sin90)^2
用sin^2+cos^2=1
同sin^2(90-X) = cos^2 X解
拆開中間數字就+到44
應該係小學 單偶數既問題? 應該點搵 係咪最大數字除2 退1格?
冇讀好多年書 好多基本野都唔識 自修緊 可否幫幫手
用YOUTUBE AD BLOCK +收二手書
白食白拎
下面答案23
44件/2 + Sin^2 90度
係44 中間呢個數字點拎到?
除2 中間數字90/2=45 ? 88/2=44?
但因為雙數跳 45唔存在
所以寫前一個數字 後一個落公式 抵消
白食白拎
下面答案23
44件/2 + Sin^2 90度
係44 中間呢個數字點拎到?
除2 中間數字90/2=45 ? 88/2=44?
但因為雙數跳 45唔存在
所以寫前一個數字 後一個落公式 抵消
點做
俾d Permutation, combination + probability 搞到好撚頭痛
(a)
Note that 7^(4n) ≡ 1 (mod 10) for n is a non-negative integer
1997^2036 ≡ 7^2036 ≡ 7^(4(509)) ≡ 1 (mod 10)
So the last digit of 1997^2036 is 1
Alternatively, consider 1997^2036 = (2000 - 3)^2036.
The first digit of 1997^2036 is the last digit of 3^2036
Note that the last digit of 3^(4n) is 1 for n is a non-negative integer
3^2036 = 3^(4(509))
So the last digit of 1997^2036 is 1
Note that 7^(4n) ≡ 1 (mod 10) for n is a non-negative integer
1997^2036 ≡ 7^2036 ≡ 7^(4(509)) ≡ 1 (mod 10)
So the last digit of 1997^2036 is 1
Alternatively, consider 1997^2036 = (2000 - 3)^2036.
The first digit of 1997^2036 is the last digit of 3^2036
Note that the last digit of 3^(4n) is 1 for n is a non-negative integer
3^2036 = 3^(4(509))
So the last digit of 1997^2036 is 1
(b)
log (1997^2036) = 2036 log 1997 = 6719.56974008
1997^2036 = (10^6719)(10^0.56974008)
The first digit of 1997^2036 is the first digit of (10^0.56974008), which is 3
log (1997^2036) = 2036 log 1997 = 6719.56974008
1997^2036 = (10^6719)(10^0.56974008)
The first digit of 1997^2036 is the first digit of (10^0.56974008), which is 3
7既POWER個位數一定係7/9/3/1/7/9/3/1...跟住就秒左
諗緊有無可能唔用計數機做
我咁做囉
原來話b
咁應該冇可能
咁應該冇可能
講緊first digit..
Number theory高手應該識
但我唔識
Number theory高手應該識
但我唔識
終於唔使zoom
屌好撚憎用zoom教書
屌好撚憎用zoom教書
推post
漏咗3(c) Find the sum of the infinite series
有冇人識做😭😭😭
試下右邊用 compound angle formula 拆
貼返喺度先，廢事又推一輪
有冇詳細步驟🙏🏼🙏🏼🙏🏼🙏🏼🙏🏼