2025年3月6日 · A problem of concept formation in which people are given the ordered triple of numbers (2, 4, 6) and are invited to try to generate further examples of triples conforming to an unspecified rule that the example obeys, trying to home in on the correct rule on the basis of simple right/wrong feedback after every guess.

If 2, 4, 6, and 9 are the digits of two 2-digit integers, what is the least possible positive difference between the integers?

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2017年10月15日 · The 2-4-6 Puzzle teaches an invaluable lesson about inductive reasoning, confirmation bias, and how “wrong” answers lead to deeper knowledge. It can be adapted with different rules for endless play options and difficulties.

2023年3月28日 · You can play my implementation of the 2-4-6 problem here (should only take a few minutes). For those of you who already know the solution but still want to test your inductive reasoning skills, I've made some more problems which work …

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2021年4月26日 · 写一个等式，左侧只能有2，4，6，6这4个数，右侧只能有41这个数，中间可以添加运算符，使得等式成立。 前者是一个算式，侧重于运算过程；后者可以理解为，是一个数，侧重于结果。 大家别想多啦，这就是小孩子的题而已。 首先加减乘阶乘根号这种运算都会得到偶数。 所以只能用除法生成奇数然后做其他运算，并且只能是6/2。 这个问题应该是无解，也可能是我算漏了. （为什么？ 写一个等式，左侧只能有2，4，6，6这4个数，右侧只能有41这个数，中 …

2+4+6一直加到100等于多少？ 两种方法1）等差数列求和，（首项+末项）X项数/2(2+100)X50/2=102X25=25502) 1+2+3+4+......+100=5050 由奇数数列和偶数数列组成奇数列1+3+5+7+.......+99偶数列2+4+6+8......+10

2016年10月21日 · 解答：2+4+6+......+2n=n（2+2n）/2=n（n+1）=n²+n. 点评：这道题考察的是等差数列的求和，只要观察其规律套用公式就可以了。 原理：等差数列是指从第二项起，每一项与它的前一项的差等于同一个常数的一种数列，常用A、P表示。 这个常数叫做等差数列的公差，公差常用字母d表示。 通项公式 为：an=a1+ (n-1)*d。 首项a1=1，公差d=2。 前n项和公式为：Sn=a1*n+ [n* (n-1)*d]/2或Sn= [n* (a1+an)]/2。 注意：以上n均属于 正整数。 扩展资料. 特殊 …