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
A Model A cylinder has radius
A Model B cylinder has radius
The height of a Model B cylinder is 25 cm.
Consider the top-down view of the open Box A shown.
We label the rectangle as
Since each circle has radius 10 cm, then each circle has diameter 20 cm, and so can be enclosed in a square with side length 20 cm whose sides are parallel to the sides of rectangle
Each circle touches each of the four sides of its enclosing square.
Because each of the circles touches one or two sides of rectangle
Therefore,
Finally, since the height of each Model A cylinder is 16 cm, then the height of the box is 16 cm, and so the volume of Box A is
The volume of Box B is equal to the volume of Box A.
Intuitively, this makes some sense since the volume of a Model B cylinder is equal to the volume of a Model A cylinder and each box is tightly packed with six cylinders.
We may follow the approach used in part (c) to demonstrate that the volume of Box B is also 38 400 cm
Recall from part (b) that a Model B cylinder has a radius of 8 cm and a height of 25 cm.
The length of Box B is six times the radius of a Model B cylinder, or
The width of Box B is four times the radius of a Model B cylinder, or
The height of Box B equals the height of a Model B cylinder, or 25 cm.
Therefore, the volume of Box B is
A quarter is worth $0.25 and so 3 quarters are worth
A dime is worth $0.10 and so 18 dimes are worth
A nickel is worth $0.05 and so 25 nickels are worth
The total value of Susan’s coins is
Solution 1
Pair each of Allen’s nickels with a dime.
Since there are an equal number of nickels and dimes (and no other coins), then every nickel pairs with a dime with no coins left over.
Each nickel
Since Allen’s coins have a total value of $1.50, then he has
Since there is one nickel in each nickel
Solution 2
Let the number of nickels that Allen has be
Allen has equal numbers of nickels and dimes and so the number of dimes Allen has is also
The value of
The value of
Since the total value of Allen’s coins is $1.50, then
Allen has 10 nickels.
A quarter is worth $0.25 and so
A dime is worth $0.10 and so
Since Elise has $10.65 in quarters and dimes, then
Using the formula given, the sum of the first 200 positive integers
The sum of the first 200 positive integers is equal to the sum of the first 150 positive integers added to the sum of the 50 consecutive integers
That is,
From part (a), the sum of the first 200 positive integers is 20 100.
Using the formula, the sum of the first 150 positive integers
Let the required sum,
Let the sum of the first 333 positive multiples of 3,
The sum of the first 1000 positive integers is equal to
That is,
Using the formula, the sum of the first 1000 positive integers
We must move the token from r1c9 to r12c4 in as few steps as possible.
At least 5
One way to move the token from r1c9 to r12c4 is to use 5
This is 8 steps in total.
Can the token be moved from r1c9 to r12c4 in fewer than 8 steps?
From above, at least 5
To move the token up the remaining 6 rows, at least 3 more steps are needed, since any step moves the token up at most 2 rows.
Therefore, at least 8 steps are required, so the minimum number of steps required is 8.
We must move the token from r1c9 to r12c4 using as many steps as possible.
Each step moves the token at least 1 row higher, so to move
The following diagram shows that this can be done using exactly 11 steps:
Note that no
Therefore, the maximum number of steps required is 11.
(Note that there are many other 11 step paths which end at
Solution 1
After 1 step, there are three different hexagons that can be reached, each in exactly 1 way. (These are the hexagons that can be reached using 1
For each positive integer
Furthermore, we can determine the number of ways that each of these new hexagons can be reached in exactly
The number of ways that a particular hexagon can be reached in exactly
This is because every path to a particular hexagon that includes exactly
For example, in the given diagram, the shaded hexagons can be reached in exactly
We also note that there are exactly
This fact allows us to verify at each step that we have accounted for all possibilities.
Using these facts, we carefully determine the hexagons that can be reached in exactly 2, 3, 4, and 5 steps, and the number of ways that each of these hexagons can be reached.
The following diagrams show this information:
From the final diagram, we see that there are exactly 6 hexagons that can be reached in each 5 steps in at least 20 ways. In other words,
Solution 2
We note that, when a fixed set of steps from a given path are rearranged, the endpoint of the path does not change.
This is because each step changes the row number and column number in a pre-determined way and so the outcome of each step is not influenced by the other steps in a path.
In other words, the paths
Since there are 3 different kinds of steps (
5 of the same step, or
4 of one step and 1 of another, or
3 of one step and 2 of another, or
3 of one step, 1 of a second type, and 1 of the third type, or
2 of one step, 2 of a second type, and 1 of the third type.
These are the only possible combinations of different types of steps.
Each specific collection of 5 steps can possibly be re-arranged in a number of different ways.
We count the number of paths in each category:
5 of the same step
There are 3 choices of the type of step.
For each choice, there is only 1 way to arrange the 5 steps.
Therefore, starting from r1c9, there is thus 1 path to each of r6c4 (5
4 of one step and 1 of another
There are 3 choices for the type of step to be taken 4 times (we call this choice
There are thus
Given a choice of
Therefore, starting from r1c9, there are thus 5 paths to each of r7c5 (
3 of one step and 2 of another
There are 3 choices for the type of step to be taken 3 times (we call this choice
There are thus
Given a choice of
1st and 2nd, 1st and 3rd, 1st and 4th, 1st and 5th,
2nd and 3rd, 2nd and 4th, 2nd and 5th,
3rd and 4th, 3rd and 5th,
4th and 5th
after which the
Therefore, starting from r1c9, there are thus 10 paths to each of r8c6 (
3 of one step, 1 of a second type, and 1 of the third type
There are 3 choices for the type of step to be taken 3 times (we call this choice
There are thus
Given a choice of
Therefore, starting from r1c9, there are thus 20 paths to each of r7c7 (
2 of one step, 2 of a second type, and 1 of the third type
There are 3 choices for the type of step to be taken 1 time (we call this choice
There are thus
Given a choice of
Therefore, starting from r1c9, there are thus 30 paths to each of r8c8 (
Therefore, in summary, there are 6 distinct hexagons (r7c7, r7c11, r9c9, r8c8, r7c9, r8c10) which are the endpoints of at least 20 paths of 5 steps; in other words,