數學小遊戲 (已結束)

mungchacha

第一題 : 3^21 (3既21次方)

第二題 : 3,125

第三題 : (冇乜心機想，谷歌找的答案，外星兄見諒)
說了真話 : D
偷吃者    : C


第四題 : C
先假設雞蛋數量：假設甲農婦有120隻雞蛋，乙農婦亦有120隻雞蛋。
如分開出售，甲農婦全部售出有60元進帳，甲農婦全部售出有40元進帳。總進帳為100元。
如合共240隻雞蛋，以2元售出5個計，總進帳只為96元。
所以答案是 C (賠)

qbwwdp

第一題 : 3

第二題 : 3125

第三題 :
說了真話 : C
偷吃者　 : B

第四題 : B) 賺, 因為農婦甲無返去問農婦乙拎錢...

張家輝

HAMAN 發表於 2012-4-28 00:56
假日前汐，有四條數學考考大家師兄弟姊妹
動動腦筋，一齊玩吓！

第一題 : 21的3次方

第二題 : 3125

第三題 :
說了真話 : D
偷吃者    : C

第四題 : C
本來兩個人分開賣的時候，雞蛋的價錢分別是1/2元和1/3元，這樣總的價格就應該是。而農婦乙按照2元錢5個賣，也就是說蛋價2/5元錢每個，這樣每個雞蛋的價錢的損失量就是:5/6 除 2 - 2/5 =1/60元錢。這樣看來，農婦乙每賣出去一個雞蛋就要損失1/60元錢，當然會少錢了。

KNVB

HAMAN 發表於 2012-4-28 00:56
假日前汐，有四條數學考考大家師兄弟姊妹
動動腦筋，一齊玩吓！

第一題 :321
第二題 :3125
第三題 :
說了真話 : D
偷吃者    :C
第四題 :C) 賠,如農婦甲&乙同樣帶來30pcs,分開賣的話,甲可得$15,乙可得$10,如混合賣只得$24

tmc2010

第一題 : 321
第二題 : 1
第三題 :
說了真話 : C   偷吃者    :  A
第四題 : A或C均可
假設農婦甲X有12 隻大雞蛋,農婦乙Y有12隻小雞蛋.此假設是基於兩者都有相同的雞蛋數以及能整數全部賣出沒有餘數(3及2的公倍數).
首先各自賣出後能得到的總價值:
X= 2隻/$1    Y=3隻/$1
Total X= 12/2=$6    Y=12/3=4  
Grand Total: $10
混合後(共24隻蛋)能得到的總價值:
(1) 2大3小/$2 ----->$8,剩4大;以2隻/$1 賣出---->$2
Total : $10
(2)3大2小/$2------>$8;剩 4小;以3隻/$1 賣出---->$1 餘1小
Total : $9 餘 1小
(3)4大1小/$2------>$6;剩9小;以3隻/$1 賣出---->$3
Total : $9
(4)4小1大/$2------>$6;剩9大;以2隻/$1賣出----->$4餘1大
Total: $10 餘1大.
如此類推:
(5)2小1大/$1------->$9
(6)2大1小/$1------->$8
(7)5小/$2----->$10餘2小
由於最後條件為乙比甲更先賣光.因此以上只有1,4,5 式成立.因此結論是A或C

HAMAN

歡樂時光過得特別快
時間到
謝謝參與

逗利是佛

逗利是佛 發表於 2012-4-29 06:23
自己笑吓自己先。。。

正死蠢答左12^3都會唔答21^3

啱啱都想打呢野....

依家真係......

懵既二次方啦

long1083

黎遲咗,冇得玩

藍天白雲

HAMAN 發表於 2012-4-28 01:08
比豬兄霸咗頭位添
留位答案
都要答才有分嫁

第二題是偽命題,因為1不等於5,只能"表示"5, 而且在數學裏,A=B並不表示B=A,如果要成立,它們之間必須是"全等"的關係才能成立.當然做為數字小遊戲則另當別論.

PrivateGump

henryyu 發表於 2012-4-30 09:41
I am not sure if my working is correct, but it would seem that 2的31次方 = 2147483648, which is bi ...

ln(2^31) = 31 * ln(2) = 31 * 0.301030 =   9.331930
ln(3^21) = 21 * ln(3) = 21 * 0.477121 = 10.019546

所以，3^21 > 2^31

用另一個方法:

2^31 = (2^30) * 2 = [(2^3)^10] * 2
3^21 = (3^20) * 3 = [(3^2)^10] * 3

單看 (2^3) = 8 & (3^2) = 9，
(2^3) ＜ (3^2)
(2^3)^10 < (3^2)^10
[(2^3)^10] * 2 < [(3^2)^10] * 3
2^31 < 3^21

oldcake

很多人都錯Q2.....包括我

HAMAN

henryyu 發表於 2012-4-30 09:41
I am not sure if my working is correct, but it would seem that 2的31次方 = 2147483648, which is bi ...

Would you use a calculator to check the answer?

HAMAN

PrivateGump 發表於 2012-4-30 10:03
ln(2^31) = 31 * ln(2) = 31 * 0.301030 =   9.331930
ln(3^21) = 21 * ln(3) = 21 * 0.477121 = 10.019 ...

Great !
but another method using calculator to know the answer directly.
Is it a better or simple ?

PrivateGump

HAMAN 發表於 2012-4-30 10:37
Great !
but another method using calculator to know the answer directly.
Is it a better or simple  ...

if you are only interested in seeing which number is larger, i have a second method above

unless if you need to know the exact number, the use of a calculator is unnecessary

HAMAN

PrivateGump 發表於 2012-4-30 10:42
if you are only interested in seeing which number is larger, i have a second method above

unless  ...

But actual no for 3^21 = 10460353200 not 10.019546
actual no. for 2^31 = 2147483648 not 9.331930

I mean calculator is a simple method to konw the answer
meanwhile, you also need a calculator to find the log no.

PrivateGump

HAMAN 發表於 2012-4-30 10:51
But actual no for 3^21 = 10460353200 not 10.019546
actual no. for 2^31 = 2147483648 not 9.331930
...

the relationship between 2 numbers X & Y, will remain after taking the log of each number.

eg. X < Y, implies ln(X) < ln(Y)

my second method above does not require a calculator to determine if 2^31 is larger than 3^21

HAMAN

藍天白雲 發表於 2012-4-30 08:00
第二題是偽命題,因為1不等於5,只能"表示"5, 而且在數學裏,A=B並不表示B=A,如果要成立,它們之間必須是"全 ...

剛剛相反，在數學上
A = B 必須是 B = A 才能成立 formula

第二題正正不是數字小遊戲
而是文字上的小遊戲，
將人對數字根深蒂固思維玩弄

HAMAN

PrivateGump 發表於 2012-4-30 11:03
the relationship between 2 numbers X & Y, will remain after taking the log of each number.

eg. X  ...

Come on
Are you not really use a calculator?

ln(2^31 ) = 31*ln(2) = A
ln(3^21) = 21*ln(3) = B

I know ln(X) > ln(Y)
but  31*ln(Y)  / 21*ln(X)
Please tell me which one is bigger?

PrivateGump

HAMAN 發表於 2012-4-30 11:15
Come on
Are you not really use a calculator?

to determine which number (2^31) or (3^21) is bigger using a calculator is slower in fact.
see my method heresee my solution here

i suppose it is convincing to say that 8<9
==> 8 = 2^3 < 9 = 3^2
==> (2^3)^10 < (3^2)^10
==> 2^30 < 3^20
==> (2^30) * 2 < (3^20) * 3, since we know that 2 < 3, and (2^30) & (3^20) are positive
==> 2^31 < 3^21
with intuition, it is faster for me to determine if 2^31 < 3^21 without calculator.
and of course, i would need a calculator to find the exact number.