As we’ve sometimes seen, the next term in a scalar sequence can be anything, since we can always find a criterion – even a mathematical algorithm – that leads to it. This is why there is a paradox that some gifted children score lower on a number sequence test because they discover more subtle than obvious ones. Let’s look at a trivial example: the next term in the sequence 1, 2, 3, 4 … is clearly 5; But it can also be 6, 7 or 8 (I invite skilled readers to find the paths that lead to these results).
Therefore, the two solutions presented by readers like the next chapter in succession put forward last week in honor of the New Year: 2000, 2002, 2020, 2022 … ie 2040 and 2200, both are correct, although the path to the second is clearer and more direct, because The next number made up of only zeros and twos. Another (crazy crazy) way to get to the same solution is to make 2 = 1, in which we will have a sequence of consecutive binary numbers: 1000, 1001, 1010, 1011, 1100…
As for the other sequence, 62, 138, 262, 446… its terms are 2022 in bases 3, 4, 5 and 6 respectively, so the next term will be 2022 in base 7, i.e. 702. Julio Diaz – Laviada came up with Same solution through a longer path:
“The second sequence appears to follow the function 2 x³ + 12x² + 26x + 22. Very easy, said Carlo, unless there is a better way… f(1), f(2), f(3), f(4) ) match, and f(5) = 702″.
Regarding the characteristics of the number 2022, perhaps most notable is its “abundance,” as Salva Foster points out:
“In terms of gluttony and gluttony, which might give way to show, comment that 2022 is plentiful, because the sum of the natural divisors excluding the number itself is 2034. May 2022 offers you a lot of positive.” Trustworthy. (Remember that a large number is a number less than the sum of its divisors without counting the number itself.) It should also be noted that 2022 is part of two Pythagorean triples (one as a leg and the other as a chord): 2022² + 2696² = 3370², 1728² + 1050² = 2022². By the way, what condition must a number meet to be part of a Pythagorean triple?
From the difference and combinations
In the comments section last week, Manuel Amoros raised an interesting combinatorial issue that led to an extensive discussion:
In the knockout cycle, two contenders appear. How many different leagues can there be? (See comments 62 to 98).
In the same assembly line, here’s an issue raised recently on university entrance exams in Turkey, which long-suffering examinees described as “diabolical” (although my savvy readers certainly wouldn’t find it that difficult):
A team of 100 people handles a certain number of projects. Everyone on the team participates in the same number of projects, and no one has the same set of projects as the other. This wouldn’t be possible if everyone would be interested in 3 projects, but it would be possible for everyone to be interested in 4. How many projects are there in total, given that there are more than 5 and no more than 10?
Carlo Frappetti Writer, mathematician, and member of the New York Academy of Sciences. He has published more than 50 popular scientific works for adults, children, and youth, including “Damn physics”, “Damn mathematics” and “The great game”. He was the screenwriter of the movie “La bola de cristal”.
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