Wednesday, April 12, 2017 (in North America and South America)
Thursday, April 13, 2017 (outside of North American and South America)
©2017 University of Waterloo
Since
Since
In
In
Since
In
Solution 1
Since
Since
Solution 2
Since
Since
We complete Row 5 of the table, as shown.
After having completed
Therefore, after having completed 5 rows,
The 25
After having completed 10 rows,
Therefore, the 100
Row
Beginning at 1, the first odd integer is
Beginning at 1, the
Row 10 ends at the
After completing 44 rows,
After completing 45 rows,
Therefore, the 2017
The final integer written in Row 45 is the 45
Each row, after the first, contains all of the integers that were written in the previous row, followed by the next two consecutive integers.
For example, Row 3 contains all integers that were written in Row 2 (that is,
Therefore, the first time an integer is written in the table, it appears as the last integer in a row, or as the second last integer in a row.
Since each row ends at the
The first time the integer 96 appears in the table, it is the second last integer written in a row (since 96 is even), and 97 is the last integer written in that same row.
Since Row
That is, Row 49 ends with the integer 97 and so the second last integer written in Row 49 is 96.
Since 96 first appears in Row 49 of the table, then 96 appears in every row following Row 49 and does not appear in any row before Row 49.
Therefore, the integer 96 appears in
When the line
Solving this equation, we get
The point with
Therefore, the line intersects the parabola at the point
The line passes through the point
The line has slope
When the line
Solving this equation, we get
The point with
Therefore, the line intersects the parabola at the point
Similarly, the line also intersects the parabola at the point
Simplifying this slope, we get
The line has slope
Therefore, the equation of the line is
(The equation of a line having slope
The
When the curve
Simplifying this equation, we get
Since
Therefore, the
Since the
That is, all points
Substituting, we get
(We may verify that
Let
We will determine
We begin by trying to construct a zigzag number with
It cannot be that
That is, the middle of the first three adjacent digits would be less than the first digit and greater than the third digit.
In our attempt to find the maximum possible zigzag number, we choose
Since
At this point, we have
We continue to assign values to digits, ensuring that for each group of three adjacent digits, either the middle digit is greater than each of the other two digits or the middle digit is less than each of the other two digits.
Since
This tells us that
Of the remaining digits, we choose
Finally, since
We proceed by proving several facts.
Fact 1:
Consider a 6-digit zigzag number counted by
We form a new 6-digit number by subtracting each of the digits of
Consider now an arbitrary 6-digit zigzag number counted by
We form a new 6-digit number by subtracting each of the digits of
Since
Since
Since
Similarly,
Therefore,
Also, if
This means that
In other words, each zigzag number counted by
However, we can apply the same process to an arbitrary zigzag number counted by
In other words,
Fact 2:
Using a similar argument to that in Fact #1, we can show that
We can now prove the statement from (ii):
We note further that we can generalize Fact 2 in a way that will be useful in (c).
Fact 3:
To see this, we use a similar argument to that from Fact 1, instead subtracting each digit of a zigzag number counted by
We now prove that
Fact 4: The number of zigzag numbers of the form
We use
Consider a 6-digit zigzag number counted by
We form a 5-digit number by deleting the first digit of
Note that
Consider now an arbitrary 6-digit zigzag number counted by
We form a new 6-digit number by deleting the first digit of
Note that we have
We then replace the digits
Since the digits of the
Therefore,
Also, if
This means that
In other words, each zigzag number counted by
However, we can apply the reverse process to an arbitrary zigzag number counted by
In other words,
Fact 5:
Using a similar argument to that in Fact #4, we can show that
We can now prove the statement from (i), noting that the second digit of a zigzag number counted by
Before concluding (b), we note the following generalization of Fact 5 that will be useful in (c):
Fact 6:
Let
Therefore,
Therefore,
We calculate the value of each of the terms on the right side by building a table of values of
Note that if
Also, the only 3-digit zigzag numbers are 132, 231, 213 and 312 because the two other ways of arranging the digits 1, 2 and 3 (123 and 321) do not satisfy the zigzag property.
Thus
1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 0 | 0 | ||||
0 | 0 | 0 | 0 | |||||
0 | 0 | 0 | ||||||
0 | 0 | |||||||
0 |
To complete the table, we build a relationship between the values of
For a positive digit
Proceeding in this way, we complete the table row by row and obtain:
1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | |
2 | 2 | 1 | 0 | 0 | 0 | 0 | 0 | |
5 | 5 | 4 | 2 | 0 | 0 | 0 | 0 | |
16 | 16 | 14 | 10 | 5 | 0 | 0 | 0 | |
61 | 61 | 56 | 46 | 32 | 16 | 0 | 0 | |
272 | 272 | 256 | 223 | 178 | 122 | 61 | 0 |
Finally,