I watched the original video. The question maker is said to be a first-year high school student. The instructor is young and very smart, but he seems to be surprised at this flexible idea of factorization. 3333 to bring it in the form of "(… 0000) ^ 2- (1) ^ 2" so that the factorization formula of a^2-b^2=(a+b)(a-b) can be used. It means that the part is tripled to 9999, but unfortunately, there are no convincing comments about the basis of the idea or the theory.
@wuchinren
2 жыл бұрын
But he SHOWED AND ONLY SHOWED that "833 and 83333 and 8333333 and 833333333 and 8333....333(8 followed even 3s in back) are all composite". Is he wrong?
@user-wu9hy4lt2w
2 жыл бұрын
@@wuchinren 様 I'm an ordinary person, so I can only understand it by multiplying it by 3 so that can be used the factorization formula. However, a more knowledgeable person who is close to a super enthusiast(otaku) or an expert may be able to explain more theoretically.
@wuchinren
2 жыл бұрын
@@user-wu9hy4lt2w Briefly (in short), 51 and 501 and 5001 and 50001 and 500001......must be divisible by 3, right?
@user-wu9hy4lt2w
2 жыл бұрын
@@wuchinren 様 The number of this problem is multiplied by 3 so that the factorization formula of a^2-b^2=(a+b)(a-b) can be used conveniently, and the square is difficult to understand, so b=1. Then, if you multiply it by 3 and add 1, it's good if the number is in the form of x^2. This problem is (83333)*3+1=249999+1=250000=(500)^2, (83333)*3=249999=250000-1=(500)^2-(1)^2=(500-1)(500+1)=499*501=(499*167)*3. Other numbers are whether they look like this. Please check for yourself.
@wuchinren
2 жыл бұрын
@@user-wu9hy4lt2w Satoh Hiroshi sama: Did you explain the problem to me? Then I may misunderstand your words before. In fact, I know how to solve the problem perfectly. So we were just trying to explain the problem to each other.
@ycyu8979
2 жыл бұрын
有意思,3连数变9连数,9连数增减1变因式分解。只要做过一次,再遇到就好办了。
@hubenbu
2 жыл бұрын
他大爺的,這麼有趣!質因數分解原型態就是相乘。
@qqqquito
2 жыл бұрын
方法奇妙!
@user-bh5pn2gn2z
2 жыл бұрын
你在教我们手动破解RSA算法?
@ugxjaaadj
2 жыл бұрын
好神奇,考试遇这种题我都是放弃的
@Celank
2 жыл бұрын
I've known this idea, but I learned this in perfectly same problem... I don't know another one by using this idea.
This kind of math is meaningless because the problem can be solved easily with a simple computer program. The following would be a better example for illustrating the power of algebra in solving problems which cannot be solved by computers using brute force. Given: a, b, c are positive integers; a/b = b/c; a + b + c = 8,191,000,273,000,003 Find: a, b, c
@walty80
2 жыл бұрын
Most problems can be solved by simple computer programs now, but that does not mean they are all meaningless. And what's the point to learn multiplication when it can be done by a $5 calculator? It's about logical thinking and insight exploration.
@fdr2275
2 жыл бұрын
@@walty80 Learning multiplication is a different story. In the old days, we were taught only the multiplication algorithm. Today, at least in the US, elementary students were taught not only the algorithm but also the theory behind the algorithm so that when they grow up and go to college, they know how to build new mathematical models. Yes, you can always argue this kind of problem can train students to think logically but what I am saying is university entrance exams should make up problems that are practical and at the same time can test students' ability of logical thinking and insightful exploration and there are plenty of problems of this kind. Problems like this one give people the impression that mathematics is a piece of joke that is used only for entertainment.
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