做题了做题了（update：是chatGPT出的一个错题。。。它怎么想出来的？）

[verdelite]

A math problem that is good for US college student:

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

[YWY]

A math problem that is good for US college student:

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

1

[verdelite]

1

改改，除1之外的最小的。

[YWY]

改改，除1之外的最小的。

2

[merrimac]

2

[verdelite]

草，这啥题目，是chatGPT出的，我轻信它了。看我们的对话过程：

我：give me an interesting math problem that is good for college student.
它：Here is an interesting math problem that might be suitable for college students:

Suppose you have a bucket that can hold exactly 3 liters of water. You also have two jugs: one that can hold exactly 5 liters of water and another that can hold exactly 8 liters of water. Using only these jugs and the bucket, how can you measure out exactly 4 liters of water?
（下面它给了解答，为了简洁我删掉了）

我：get me another one.
它：Sure! Here is another math problem that might be suitable for college students:

A farmer has 1000 meters of fencing and wants to enclose a rectangular field. He does not need to fence the entire perimeter of the field, as one side is bordered by a river. Find the dimensions of the rectangle that will enclose the maximum possible area.
（下面它给出答案，为了简洁我删掉了）

我：this one is too easy. get a more difficult one
它：Certainly! Here is a more challenging math problem that might be suitable for college students:

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

This problem involves the use of number theory and problem-solving skills. It requires the use of some basic principles from arithmetic, such as the formula for the sum of an arithmetic series, as well as some creative thinking to devise a plan for solving the problem.

Here is one way to solve the problem:

First, let's consider the elements of S that are less than 50. These can be found by adding together different sets of consecutive positive integers until the sum exceeds 50. Here is a list of all the elements of S that are less than 50:

1 = 1
3 = 1 + 2
6 = 1 + 2 + 3
10 = 1 + 2 + 3 + 4
15 = 1 + 2 + 3 + 4 + 5
21 = 1 + 2 + 3 + 4 + 5 + 6
28 = 1 + 2 + 3 + 4 + 5 + 6 + 7
36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8
45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9

We can see that all of the integers from 1 to 45 can be expressed as the sum of two or more consecutive positive integers. Therefore, the smallest positive integer that is not an element of S is 46.

To prove that 46 is indeed the smallest positive integer that is not an element of S, we need to show that 46 cannot be expressed as the sum of two or more consecutive positive integers. To do this, we can use the formula for the sum of an arithmetic series:

(n/2)(2a + (n - 1)d) = 46

Where n is the number of terms in the series, a is the first term, and d is the common difference.

We can see that if n is odd, then the sum of the series will be odd. If n is even, then the sum of the series will be even. Therefore, the only solution to this equation will be n = 1, which means that 46 cannot be expressed as the sum of two or more consecutive positive integers. This means that 46 is indeed the smallest positive integer that is not an element of S.

草，原来是个错题。它怎么想出来的？它有怎样的一个思维过程？

[YWY]

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

问：是不是有无穷多正整数不在集合S里？

[YWY]

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

问：是不是有无穷多正整数不在集合S里？

证明：一个正整数n不在集合S里当且仅当n是2的方幂（n = 2^i for some non-negative integer i）。

[resso]

A math problem that is good for US college student:

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

你把题出错了，lol，你真的有学习诶

[verdelite]

你把题出错了，lol，你真的有学习诶

这就有课题了：chatGPT 是如何思考出这样一个错题的？从中我们能总结出它的思维有什么特点，怎样能山寨出来？

[verdelite]

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

问：是不是有无穷多正整数不在集合S里？

先解决一小部分：取一个数，如果它能被一个奇数2k+1整除，则它可以被表示为2k+1个连续数之和。例如15能被3整除，则15=4+5+6。还有，18=5+6+7。

如果它不能被所有任何奇数整除，则它要么是一个质数，要么是2的幂。

看看质数，似乎不能被分解。所以有无穷多个数在S中。

(Update:原先针对质数写错了；质数或者奇数2k+1能分解，可以表示为k+(k+1)所以把“不”字划掉)。

[TheMatrix]

先解决一小部分：取一个数，如果它能被一个奇数2k+1整除，则它可以被表示为2k+1个连续数之和。例如15能被3整除，则15=4+5+6。还有，18=5+6+7。

如果它不能被所有任何奇数整除，则它要么是一个质数，要么是2的幂。

看看质数，似乎不能被分解。所以有无穷多个数在S中。

这个解的不错啊。

[YWY]

先解决一小部分：取一个数，如果它能被一个奇数2k+1整除，则它可以被表示为2k+1个连续数之和。例如15能被3整除，则15=4+5+6。还有，18=5+6+7。

如果它不能被所有任何奇数整除，则它要么是一个质数，要么是2的幂。

看看质数，似乎不能被分解。所以有无穷多个数在S中。

显然S有无限个数，1+2， 1+2+3， 1+2+3+4， 等等都在S里。另外，所有大于1的奇数都在S里，2k+1 = k + (k+1)。

我出的问题是有无限个正整数不在S中。不过我楼上给出了简单说明（没加细节）有错误。

[YWY]

Let S be the set of all positive integers that can be expressed as the sum of two or more consecutive positive integers. For example, 15 can be expressed as the sum of 3 + 4 + 5, so 15 is an element of S. Find the smallest positive integer that is not an element of S.

问：是不是有无穷多正整数不在集合S里？

先解决一小部分：取一个数，如果它能被一个奇数2k+1整除，则它可以被表示为2k+1个连续数之和。例如15能被3整除，则15=4+5+6。还有，18=5+6+7。

如果它不能被所有任何奇数整除，则它要么是一个质数，要么是2的幂。

看看质数，似乎不能被分解。所以有无穷多个数在S中。

受世人皆傻帖子的启发，感觉下面的命题是对的，大家可试着证一下（或给反例）。

命题：一个正整数n不在集合S里当且仅当n是2的方幂（n is not in S if and only if n = 2^i for some non-negative integer i）。

[denizen]

stackoverflow把它办了

其实它会个屁的思考，就是人云亦云

这道题八成还是从stackoverflow上抄的，它转写了一下而已。

要不原题就错了，要不就是它转写错了。

[verdelite]

stackoverflow把它办了

其实它会个屁的思考，就是人云亦云

这道题八成还是从stackoverflow上抄的，它转写了一下而已。

要不原题就错了，要不就是它转写错了。

你问它几个问题看看。它每次都能精准理解你的要求，你还想怎么样？人也不过如此而已。你以为人的思考是个啥，就是个屁。