Brain Teaser (1) 腦筋急轉彎(一)
During my time at uni, I made friends with a lot of physicists and mathematicians. Most of us had a common feature: we like to think about interesting problems. I guess that's the primary reason why we could survive the degree! One of them went as follows:
Suppose we have 100 light bulbs, labelled as 1,2,3....,100. Each of them comes with an on/off button so that if it's on and we press the button, it turns off and vice verse. Initially, all these light bulbs are off. Now, we press the buttons of those that are labelled by a multiple of 1. The result is that all the light bulbs are now on, because all integers are multiples of 1. Next, we press those that are labelled by a multiple of 2. Then, all the even-numbered ones are off and all the odd-numbered ones are on. We repeat this for multiples of 3, multiples of 4, etc, up to multiples of 100. The question is: Which light bulbs remain on in the end?
For those who can program, this problem is easily solved computationally, but that will not be fun because you won't have to do any thinking. If you insist to solve it computationally, change the number from 100 to 100 million million to make it more challenging!
在我上大學的時候,我交了不少修物理跟數學的朋友。事實上,我大部分的朋友都是修數理科的,大概是因為我們有一個共通點,就是我們都喜歡動腦筋,想一些有趣問題。今天想在這裏跟大家分享一下其中一個。那個問題是這樣的:
我們有一百個燈泡。每個燈泡自己有一個編號,就叫它們燈泡一、燈泡二、燈泡三,如此類推吧。每個燈泡有一個開關鈕。按一下的話,要是本來開的就會關,關的就會開,就像你家裏的燈泡一樣。開始的時候,所有的燈泡都是關掉的。現在我們把一的倍數全部按一下。結果就是所有燈泡都亮了對吧,因為所有整數都是一的倍數。然後我們把二的倍數的全部按一下:現在雙數的都關了,單數的還開着。我們為三、四、五。。。。一百的倍數重覆以上動作。到最後,哪些燈泡還是亮着的?
會編碼的朋友很容易就可以把這個問題解決了,但是那樣的話不是很好玩。
这道题我小学时候做过,奥数题啊😄
噢原來是奧數的題哈哈 其實會乘法的話就可以把這道題做出來了。
我曾经用这道题考我姐姐,她和姐夫两个人找了个房间做了很久,没做出来🤣
哈哈 有嘗試也是好事 :)
腦筋急轉彎,冬天到了為什麼大雁要往南飛
該不會是平方數吧!
這應該跟如何拆解有關,用8和9做比較好了:
我們可以把8拆解成如下形式:
1×8
2×4
4×2
8×1
共有四組
已知一開始燈是暗的,在經過偶數次的開關後仍是熄滅
而9可拆解成:
1×9
3×3
9×1
共有3組
由於是奇數次,所以燈會亮
其他也可如此類推,平方數必然只能拆解成奇數次,它則不然,為偶數次。
答案就出來了,不知這想法對不對