Question

5. In math class, a student has written down a sequence of 16 numbers on the blackboard. Below each number, a second student writes down how many times that number occurs in the se‐ quence. This results in a second sequence of 16 numbers. Below each number of the second se‐ quence, a third student writes down how many times that number occurs in the second se‐ quence. This results in a third sequence of numbers. In the same way, a fourth, fifth, sixth, and seventh student each construct a sequence from the previous one. Afterward, it turns out that the first six sequences are all different. The seventh sequence, however, turns out to be equal to the sixth sequence.

   Give one sequence that could have been the sequence written down by the first student. Explain which solution strategy or algorithm you have used.

Answer 1

Pls, give thumbs up.

The solution to this question can be obtained in many ways by considering different possible sequences.

One possible sequence : 9 9 9 9 5 8 3 3 3 3 7 7 9 9 9 9

First Student : 9 9 9 9 5 8 3 3 3 3 7 7 9 9 9 9

Second Student : 8 8 8 8 1 1 4 4 4 4 2 2 8 8   8 8

Third Student : 8 8 8 8 2 2 4 4 4 4 2 2 8 8 8 8

Fourth Student : 8 8 8 8 4 4 4 4 4 4 4 4 8 8 8 8

Fifth Student : 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

Sixth Student : 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16

Seventh Student :16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16

Observation :

The seventh sequence and sixth sequence both are same.

Strategy :

As I told there are many sequences which can yield as last two sequences same.

The sequence should be considered in such a manner that the second sequence should include twice the number 1, twice the number 2, four times the number 4 and eight times the number 8.

Even if you change alternate any value of it you will get a new sequence. ( Example : 9 9 9 5 9 8 3 3 3 3 7 7 9 9 9 9 )