Thursday, April 16, 2015 (in North America and South America)
Friday, April 17, 2015 (outside of North American and South America)
©2015 University of Waterloo
The distance from the end of any car (boxcar or engine car) to the end of the next car is the sum of the 2 m distance between the cars and the 15 m length of a boxcar, or 17 m.
Thus the distance from the end of the engine car to the end of the 10
The total length of a train with
That is, the total length of a train with
If a train has a length of 2015 m, then
A train with 14 boxcars has length
The length of time during which a portion of the train is in Canada and a portion of the train is in the United States at the same time is equal to the total length of time it takes the train to cross the border. (When the train is crossing the border, portions of the train are in both countries at the same time.)
The train begins to cross the border when the front of the engine car reaches the border. The train finishes crossing the border when the end of the last boxcar reaches the border and so the front is 264 m farther.
That is, the length of time required for the train to cross the border is equal to the length of time it takes the train to travel a distance equal to the length of the train, or 264 m.
Since the train is travelling at a speed of 1.6 m/s, then the time required to travel 264 m is
The two-digit positive integers
Solving
Since
Therefore, the positive integer
The two-digit positive integers
Solving
Since
However, 80 is not a multiple of 9 and so
The three-digit positive integers
Simplifying
Since
Since
That is,
(Verify for yourself that there are values for
Since
Diagram 1 illustrates that
To determine
We are told that this new (4th) line segment must intersect each of the existing 3 line segments exactly once, creating 3 new points of intersection (labelled
This 4
In addition, each of the points which exist in the illustration of
Therefore, we get
To determine
We are told that this new (5th) line segment must intersect each of the existing 4 line segments exactly once, creating 4 new points of intersection (labelled
This 5
In addition, each of the points which exist in the illustration of
Therefore, we get
Therefore,
As in part (a), consider finding
To determine
This new (
This
In addition, each of the points which exist in the illustration of
From part (b),
That is, the addition of an
For example since
For small values of
2 | |
3 | |
4 | |
5 | |
6 | |
⋮ | ⋮ |
We may use the pattern in the table above to establish an equation for
What is the pattern?
Consider for example the row for
This is true for each of the rows shown in the table.
That is,
(Verify that this is true for each of the rows shown in the table.)
Since the addition of an
Reorganizing this equation for
Solving
Since
Therefore, for each positive integer
The
These
The possible values of
There are 25 multiples of 5 between 1 and 125, inclusive.
5 of these have a
The remaining
Since there are 25 multiples of
1 integer
4 integers
20 integers
100 integers
Therefore,
Since
Since
Since
Therefore,
To calculate
For each positive integer
The only multiple of
There are thus
Therefore,
Therefore,
We prove that
From (b), we obtain
Since the right side is a power of a prime number, then the only divisors of the right side are powers of
Therefore, each of the factors on the left side must be powers of
Since
Therefore,
If
If
This also tells us that if
Can
If
If
Thus, if
Therefore, if
If
But in this case,
This contradicts our original assumption.
Therefore,
We note that
Suppose that
Then
Are there prime numbers
Note that if
Therefore, we could have
Therefore,
Since
Let’s next see if
Using a similar derivation to that in (a), we can see that
If
Therefore,
Thus,