As we have seen on occasion, the next term in a numerical sequence can be anything, since we can always find a criterion -even a mathematical algorithm- that leads to it. That is why there is the paradox that some gifted children obtain low scores on the number sequence test because they discover more subtle relationships than the obvious ones. Let’s look at a trivial example: the next term in the sequence 1, 2, 3, 4… is obviously 5; but it could also be 6, 7 or 8 (I invite my astute readers to find the paths that lead to these results).
Therefore, the two solutions provided by readers as the next term in the succession raised last week in honor of the new year: 2000, 2002, 2020, 2022 …, namely 2040 and 2200, are both valid, although the path leading to the second is clearer and more direct, since it is the next number made up of zeroes and twos only. Another (crazily crazy) way to reach the same solution is to make 2 = 1, with which we would have a sequence of successive 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, that is, 702. Julio Díaz-Laviada reached the same solution by a longer path:
“The second sequence seems to follow the function 2 x³ + 12x² + 26x + 22. Very easy, as Carlo said, unless there is some better method… f (1), f (2), f (3) and f (4) match, and f (5) = 702 ″.
As regards the properties of the number 2022, perhaps the most remarkable is its “abundance”, as Salva Fuster points out:
“Related to voracity and gluttony, which may give way to width, comment that 2022 is abundant, since the sum of its natural divisors except the number itself is 2034. May 2022 offer you an abundance of positive things.” Amen. (Remember that an abundant number is one that is less than the sum of its divisors without counting the number itself). And it should also be noted that 2022 is part of two Pythagorean triples (one as a leg and the other as a hypotenuse): 2022² + 2696² = 3370², 1728² + 1050² = 2022². By the way, what condition must a number meet in order to be part of a Pythagorean triple?
Of teams and combinations
In last week’s comment section, Manuel Amorós raised an interesting combinatorial problem that led to extensive discussion:
In a knockout tournament, 2n competitors appear. How many different tournaments can there be? (See comments from 62 to 98).
In the same combinatorial line, here is a problem recently raised in the university entrance exams in Turkey, which was described as “devilish” by the long-suffering examinees (although surely my astute readers will not find it so difficult) :
A team of 100 people tackles a certain number of projects. Everyone on the team participates in the same number of projects, and no one has the same combination of projects as another. This would not be possible if each person took care of 3 projects, but it is possible if each one takes care of 4. How many projects are there in total, knowing that there are more than 5 and not more than 10?
Carlo Frabetti is a writer and mathematician, member of the New York Academy of Sciences. He has published more than 50 popular science works for adults, children and young people, including ‘Damn physics’, ‘Damn mathematics’ or ‘The great game’. He was a screenwriter for ‘La bola de cristal’.
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