Author Topic: ZT: 2、 3 、10、 10算24点,你能算出来吗?  (Read 9263 times)

万精油

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ZT: 2、 3 、10、 10算24点,你能算出来吗?
« on: 八月 08, 2011, 11:17:04 am »
从开心一刻移到这里,因为话题开始向做题方向移动。:)

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我觉得这不是很好笑,但因为与数字有关,可以加分。:), 所以转载一下。

--万精油--

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这尼玛是人类算24点的方法!!!!
事件缘由:今天晚上突然看到一条24点状态,当时惊为天人,这NM叫人啊?以下是那条
状态

朱明西 : 24点,算2 3 10 10,我LX炮狗等面对四张牌痛不欲生,结果跑跑同学扫了一
眼说,算出来了,2的10次方减10的3次方。。我草这是人类的算24点啊。。

然后么。。。我就在深夜很得瑟的问室友求室友算

刚出完题,文哥的暴走之旅开始了

5秒后,解法一出炉

2*(10-3)+10

好吧 我傻逼 这么简单的算法我居然不知道

但是我不服气啊!然后我就逼着文哥换一种方法

又是5秒钟。。。。。。。解法二诞生

2*10+3+log10

我阵亡了。。。真的 然后继续要求文哥换方法

这次稍微久一点。。。。大概10秒钟吧

解法三: 2^10-10^3

我彻底折服了、、、文哥在总思考时间约30秒的情况下算出了3种解法。。。

这样就结束了吗?你错了!!

在文哥N分钟后爆出虚数好像不行的语句后,我们全寝室开始鼓励文哥给力

然后,

解法四:(2+3-10/10)!

全寝震精了。。。阶乘都出来了。。。。。。这是人类在算24点么

在文哥销声匿迹一段时间后。。。我开始群嘲了。。。开始挑战文哥的权威,跟他打赌
他找不到新的方法,可是事实证明我错了,错的是那么离谱,我败得是如此的体无完肤

好吧,直接列出解法五

(10C3)*2/10

下面是最非人类的解法!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
通解公式:((A’)!+(B’)!+(C’)!+(D’)!)!

经过逼友们的建议,正式命名该式子为马景涛通解公式!!!!!咆哮吧,我的24点!
!!!!!!!

以后算24点无敌啦!!西西!!!!!
还有其他BT解法。。

受教啊……3*(10-2)*log10

呃……一不小心第六种了。。。
(根号下10X10 —2)X3

反正试试又不花钱
第七种
(10+2)÷sin(3*10)

(10-2)*min(3,10) 噗第八
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« Last Edit: 八月 20, 2011, 12:01:03 pm by 万精油 »

万精油

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Re: ZT: 2、 3 、10、 10算24点,你能算出来吗?
« Reply #1 on: 八月 08, 2011, 11:23:17 am »
发帖后突然想到象这样什么运算都可以用的话,可以轻松搞出很多算法

10+10+3!-2

(min(10,10)-2)*3

(max(10,10)-2)*3

(10+2)/cos(3!*10)

10+10+3+(2->)

(10-(3<-))+2*10

....

packman

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Re: ZT: 2、 3 、10、 10算24点,你能算出来吗?
« Reply #2 on: 八月 08, 2011, 11:05:09 pm »
My interpretations:

Min and max? Come on, that is cheating. :wink: You can ignore the rest numbers if you get 24 with only two or three numbers.
Trigs? Sin(3*10)? what is the unit of 3 or 10? Sqrt of degree?
Exponential is fine, just like multiplication. 2 * 3 is 2 added 3 times, as to 2^3 is 2 multiplied 3 times.
In the same token, square root  or other roots can be used only if it is expressed as ^(1/n). Log can be used only if the base is written out.
Factorial? No. It introduces extra numbers (5! introduces 1 thru 4).
简单==完美

fzy

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Re: ZT: 2、 3 、10、 10算24点,你能算出来吗?
« Reply #3 on: 八月 09, 2011, 02:18:43 pm »
This is an earlier IBM problem: Using two 2's and any common mathematical functions and operations, to obtain 5.

After some thoughts, I was able to extend it to this: Given two arbitrary rational numbers r1 and r2, using only the functions available on a calculator, ie, exp, ln, sin, cos, tan, arcsin, arccos, arctan, and - (unary negation), start from r1 and end at r2. For example, exp(0) = 1.

But as usual, I don't seem to remember how that was done.

万精油

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Re: ZT: 2、 3 、10、 10算24点,你能算出来吗?
« Reply #4 on: 八月 09, 2011, 02:40:28 pm »
Quote
Given two arbitrary rational numbers r1 and r2, using only the functions available on a calculator,
ie, exp, ln, sin, cos, tan, arcsin, arccos, arctan, and - (unary negation), start from r1 and end at r2.
For example, exp(0) = 1.


Most calculators have the sqrt operator (e.g. the one that comes with Windows has sqrt).

Starting from any number, push sqrt enough times, you will get 1.

Most calculators have the feature that if you push +, the current number will be stored as the number
to be added, then push =, the sum of the current number and the stored number will be showing.
Keep push the = operator, the stored number will be added again and again. Thus, starting from 1,
you can get any number. :)


万精油

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Re: ZT: 2、 3 、10、 10算24点,你能算出来吗?
« Reply #5 on: 八月 09, 2011, 02:43:07 pm »
As a side note, a similar problem is, given four 4's, using the common operator (+,-,*,/, !, ^, sqrt), we can get every number from 1 to 72, I haven't been able to get 73.

If we have more discussion on this, I will move this to the puzzle forum. :)

fzy

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Re: ZT: 2、 3 、10、 10算24点,你能算出来吗?
« Reply #6 on: 八月 10, 2011, 04:55:28 pm »
Oh please! This is the last forum where most people are not stupid. I really wished I do not need to state all the necessary conditions as clearly as possible.

It is specifically stated in the IBM problem that using square root counted as using a 2, as well as using base 2 log.

Also the two sides need to be mathematically equal.

万精油

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Re: ZT: 2、 3 、10、 10算24点,你能算出来吗?
« Reply #7 on: 八月 20, 2011, 12:08:12 pm »
I almost missed this reply from fzy (due to the forum was having trouble for two days, and the reply was during those two days or immediately before it), just found out it today. And I think it is moving towards puzzle, so I moved the topic to this  forum.

Quote
Oh please! This is the last forum where most people are not stupid. I really wished I do not need to state all the necessary conditions as clearly as possible.

Since the topic was in humor, I gave a humorous answer. Obviously, it was not that funny. :)

Quote
It is specifically stated in the IBM problem that using square root counted as using a 2, as well as using base 2 log.

I don't remember seeing this problem in IBM problems. Now that we have moved this problem to here, can you gave a link of the IBM problem so that we know what is exactly the requirement (what are allowed and what are not). Thanks.


Dr Kevin Wang

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Re: ZT: 2、 3 、10、 10算24点,你能算出来吗?
« Reply #8 on: 八月 23, 2011, 02:08:11 pm »
As a side note, a similar problem is, given four 4's, using the common operator (+,-,*,/, !, ^, sqrt), we can get every number from 1 to 72, I haven't been able to get 73.
If the following is allowed: decimal point in front (.4), and repeat decimal (.4~ meaning .444444444....), then 73 is possible.

If log is allowed, every positive integer is possible:
n = -log_{sqrt{4}} log_{sqrt{4}} sqrt{sqrt{sqrt{..........sqrt{4*4}....}}}
where there are n+2 nested square roots.