題目：4個人的年齡是4個連續數，他的積是120
咁他的和是？
請寫出計法。
their ages are x, x+1, x+2 and x+3
So, x (x+1) (x+2) (x+3) = 120, i.e. x^4+6x^3+11x^2+6x = 120
since x is an integer and if x = 4, x^4 = 256 > 120, then x must be < 4.
Putting x = 3, 3(4)(5)(6) = 360 > 120, and you will find out x = 2 and their ages are 2, 3, 4, 5.
Their sum is then 14
Quote from: hangchoi on 24 March 2012, 18:28:54
their ages are x, x+1, x+2 and x+3
So, x (x+1) (x+2) (x+3) = 120, i.e. x^4+6x^3+11x^2+6x = 120
since x is an integer and if x = 4, x^4 = 256 > 120, then x must be < 4.
Putting x = 3, 3(4)(5)(6) = 360 > 120, and you will find out x = 2 and their ages are 2, 3, 4, 5.
Their sum is then 14
Very good!
I missed the 2nd step about X^4=256>120 part.
我細妹話係某小六升中一面試題目
重有另外兩條
題目2：有10位同學在活動中每人向其他9人握手，10個人共互相握手多少次？
問題3: 有四個盒，每個盒子裝不少於兩個波，任何三個盒加起來不少於八個波，這四個盒至少有多少個波？
Quote from: chanchiwai on 25 March 2012, 21:06:45
題目2：有10位同學在活動中每人向其他9人握手，10個人共互相握手多少次？
10C9 = 45 ??
Quote from: chanchiwai on 25 March 2012, 21:10:33
問題3: 有四個盒，每個盒子裝不少於兩個波，任何三個盒加起來不少於八個波，這四個盒至少有多少個波？
11
Quote from: chanchiwai on 25 March 2012, 21:10:33
問題3: 有四個盒，每個盒子裝不少於兩個波，任何三個盒加起來不少於八個波，這四個盒至少有多少個波？
3 + 3 + 3 + 2 . . . .= 11
Quote from: chanchiwai on 25 March 2012, 21:06:45
題目2：有10位同學在活動中每人向其他9人握手，10個人共互相握手多少次？
小學時的計算方法：
9+8+7+6+5+4+3+2+1 = 45 ;D
Quote from: hangchoi on 25 March 2012, 21:15:53
10C9 = 45 ??
10C2!!!!Two pupils shake hands at a time!!!
Quote from: wongyan on 26 March 2012, 10:56:18
10C2!!!!Two pupils shake hands at a time!!!
Yes...make the conversion too quick.... ;D
Quote from: hangchoi on 27 March 2012, 10:27:05
Yes...make the conversion too quick.... ;D
toooooo primary school . . . that's why !!! ;D
Quote from: oo8 on 27 March 2012, 14:08:13
toooooo primary school . . . that's why !!! ;D
are you sure they are really primary school level.....i have douobt.... :o :o :o
Quote from: chanchiwai on 28 March 2012, 15:46:50
are you sure they are really primary school level.....i have douobt.... :o :o :o
We had this kind of Math question for 升中試！ >:(
But with 5 or 6 student instead of 10.