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
Red pens are sold in packages of 6 pens.
Therefore, 5 packages of red pens contain
Blue pens are sold in packages of 9 pens.
Therefore, 3 packages of blue pens contain
Altogether, Igor bought
Robin bought 21 packages of red pens which contain
Of the 369 pens that Robin bought, the number of blue pens was
Blue pens are sold in packages of 9, so the number of packages of blue pens bought by Robin was
Solution 1:
Let the number of packages of red pens bought by Susan be
Let the number of packages of blue pens bought by Susan be
Thus, Susan buys
If Susan bought exactly 31 pens, then
Factoring the left side of this equation, we get
Since
Since the right side, 31, is not a multiple of 3, there is no solution to the equation
Therefore, it is not possible for Susan to buy exactly 31 pens.
Solution 2:
Let the number of packages of red pens bought by Susan be
Let the number of packages of blue pens bought by Susan be
Thus, Susan buys
If Susan bought exactly 31 pens, then
The smallest possible value for
Solving the equation
In the table below, we use this equation to determine the value of
0 | |
1 | |
2 | |
3 |
For each of the possible values for
Thus, there is no solution to the equation
Expressing
We require that
The only integer
Expressing
Since
Expressing
Since
The integer values of
At the start of the weekend, Fiona has played 30 games and has
Fiona’s win ratio at the start of the weekend is greater than
Since
During the weekend Fiona plays five games giving her a total of
Since she wins three of these games, she now has
Fiona’s win ratio at the end of the weekend is less than
Rewriting this inequality with a common denominator of 70, we get
Since
The integer values of
Chords
Solving this equation for
The length of
Chords
Solving this equation, we get
Chords
Since
Substituting values into
Chords
Since
Substituting values into
Substituting
Therefore,
Dave, Yona and Tam have 6, 4 and 8 candies, respectively.
Since they each have an even number of candies, then no candies are discarded.
During Step 2, Dave gives half of his 6 candies to Yona and accepts half of Tam’s 8 candies so that he now has
Similarly, Yona gives half of her 4 candies to Tam and accepts half of Dave’s 6 candies so that she now has
Tam gives half of his 8 candies to Dave and accepts half of Yona’s 4 candies so that he now has
Since Dave has 7 candies, and Yona has 5 candies, they each discard one candy while Tam, who has an even number of candies, does nothing.
These next two steps are summarized in the table to the right.
Following the given procedure, we continue the table until the procedure ends, as shown.
When the procedure ends, Dave, Yona and Tam each have 4 candies.
Dave | Yona | Tam | |
---|---|---|---|
Start | 3 | 7 | 10 |
After Step 1 | 2 | 6 | 10 |
After Step 2 | 6 | 4 | 8 |
After Step 2 | 7 | 5 | 6 |
After Step 1 | 6 | 4 | 6 |
After Step 2 | 6 | 5 | 5 |
After Step 1 | 6 | 4 | 4 |
After Step 2 | 5 | 4 | 4 |
After Step 1 | 4 | 4 | 4 |
Dave, Yona and Tam begin with 16, 0 and 0 candies, respectively.
The results of each step of the procedure are shown in the table. (We ignore Step 1 when each of the students has an even number of candies.)
Each student has 4 candies when the procedure ends.
Dave | Yona | Tam | |
---|---|---|---|
Start | 16 | 0 | 0 |
After Step 2 | 8 | 8 | 0 |
After Step 2 | 4 | 8 | 4 |
After Step 2 | 4 | 6 | 6 |
After Step 2 | 5 | 5 | 6 |
After Step 1 | 4 | 4 | 6 |
After Step 2 | 5 | 4 | 5 |
After Step 1 | 4 | 4 | 4 |
We begin by investigating the result that Step 2 has on a student’s number of candies.
Assume Yona has
Further, assume that
During Step 2, Yona will give half of her candies away, leaving her with
In this same Step 2, Yona will also receive
Therefore, Yona completes Step 2 with
On Wednesday, Dave starts with
Since
That is, we begin the procedure by performing Step 1 which leaves Dave with
After Step 2, Yona will have the average of her number of candies,
Tam will have the average of his number of candies,
Dave will have the average of his number of candies,
Since Yona and Dave each now have an odd number of candies, Step 1 is performed.
The procedure is continued in the table shown.
Dave | Yona | Tam | |
---|---|---|---|
Start | |
||
After Step 1 | |||
After Step 2 | |||
After Step 1 | |||
After Step 2 | |||
After Step 1 |
At the end of the procedure, each student has
On Thursday, Dave begins with
In the table shown, we proceed with the first few steps to get a sense of what is happening early in the procedure. (We again ignore Step 1 when each student has an even number of candies.)
Dave | Yona | Tam | |
---|---|---|---|
Start | 0 | 0 | |
After Step 2 | 0 | ||
After Step 2 | |||
After Step 2 | |||
After Step 2 |
As was demonstrated in part (c), each application of Step 2 gives the average number of candies that two students had prior to the step.
If each of the three students has a number of candies that is divisible by
If Yona has
Since
We proceed by introducing 4 important facts which will lead us to our conclusion.
Important Fact #1:
We are starting with
The first application of Step 2 gives three numbers, each of which is divisible by
The second application of Step 2 gives three numbers, each of which is divisible by
The third application of Step 2 gives three numbers, each of which is divisible by
That is, starting with
We note that at each of these 2017 steps, the number of candies that each student has is even, and therefore Step 1 is never applied (no candies have been discarded), and so the total number of candies shared by the three students is still
Important Fact #2:
If we begin Step 2 with
That is, there are still exactly two students who have an equal number of candies.
Important Fact #3:
If we begin Step 2 with
Applying Step 2 to
Similarly, applying Step 2 again to this “2 high, 1 low state” gives a “2 low, 1 high state”.
Since we begin with
Important Fact #4:
Beginning with
After applying Step 2, we have
That is, applying Step 2 once decreases the positive difference between the high and low numbers of candies by a factor of 2 (that is,
Therefore, beginning with
That is, after applying Step 2 2017 times, the number of candies is
Conclusion:
Since we haven’t applied Step 1, then there are still
If
Since
Since
Two candies were discarded in the application of Step 1 and so there are now
Since each student has an equal number of candies, and there are