a) Prove that in the table
| | | | | 1 | | | | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | | | 1 | 1 | 1 | | | |
| | | 1 | 2 | 3 | 2 | 1 | | |
| | 1 | 3 | 6 | 7 | 6 | 3 | 1 | |
| | | | | | | | | |
where each number is equal to the sum of the three numbers above it, there is an even number in each row (starting from the third).
b) Is there a number divisible by 3 in every row (except the first two)?
a) In our table, we will write 0 instead of even numbers and 1 instead of odd numbers. Then, the table of zeros and ones should be constructed according to the same rule (each number is the sum of the three numbers above it in the previous row), but the addition should be performed modulo two.
The first method. The last four numbers in each row depend only on the last four numbers in the previous row, so this quartet repeats periodically (with a period of 4, see Fig. 1).
!
Fig. 1
The second method. Suppose that some row consists entirely of ones. Then the previous row, as it is easy to see, can only be: , and the one before it can only be:
110000110000...000011000011. But this is impossible, since in each row of our table there is an odd number of numbers.
b) As it is easy to see, already in the next, fourth row of the table (it is convenient to assign the number 0 to the top row consisting of one one) 1,4,10,16,19,16,10,4,1 no number is divisible by 3.
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