Wednesday, November 17, 2021
(in North America and South America)
Thursday, November 18, 2021
(outside of North American and South America)
©2021 University of Waterloo
The area of a rectangle is equal to its length
multiplied by its width, and so width equals area divided by
length.
The width of Rectangle A is equal to
The width of Rectangle B is equal to
The width of Rectangle C is equal to
The smallest of these widths is 3 cm, and so
(Note that since the three rectangles have the same area, the smallest
width would be the one with the largest length.)
Answer:
The prime numbers that are greater than 10 and less
than 20 are 11, 13, 17, 19.
(Since
The sum of these prime numbers is
Answer: 60
From the 22nd floor to the
Since Taya goes from each floor to the next in 15 seconds, this takes
her
Since Jenna waits for 2 minutes, she waits for 120 seconds. Since Jenna
goes from each floor to the next in 3 seconds, it takes a total of
Since their travel times are equal, then
Answer:
Since
Since
The sum of the measures of the angles in
Answer: 210
Using the given rules, the first 7 numbers in the
list are
In other words, the list will repeat every 5 numbers since each group of
5 numbers is produced in exactly the same way as the previous 5
numbers.
The sum of the first 5 numbers is
Since
In fact, since
We make a chart of the next terms and the sum to that point until we
reach a sum that is greater than 2021 and is an odd integer:
Term # | |||||||
---|---|---|---|---|---|---|---|
Term | |||||||
Sum to this point |
Term # | |||||||
---|---|---|---|---|---|---|---|
Term | |||||||
Sum to this point |
Therefore, the smallest positive integer
Answer: 989
Solution 1
Suppose that Dragomir removes the socks one at a time. There are 12
socks that he can remove 1st, then 11 socks 2nd, then 10 socks 3rd, and
finally 9 socks 4th.
This means that there are
Suppose that there is exactly one pair among the 4 socks removed. There
are 6 different possibilities for how this pair was formed: 1st and 2nd
socks match, 1st and 3rd match, 1st and 4th match, 2nd and 3rd match,
2nd and 4th match, 3rd and 4th match.
We determine the number of ways in which each of these possibilities can
arise.
Case 1: Matching pair 1st and 2nd
There are 12 socks that he can pick 1st (no restrictions).
There is 1 sock that he can pick 2nd (pair of 1st).
There are 10 socks that he can pick 3rd (any of the 10 remaining
socks).
There are 8 socks that he can pick 4th (any of the 9 remaining socks
other than pair of 3rd).
In this case, there are
Case 2: Matching pair 1st and 3rd
There are 12 socks that he can pick 1st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There is 1 sock that he can pick 3rd (pair of 1st).
There are 8 socks that he can pick 4th (any other than pair of
2nd).
In this case, there are
Case 3: Matching pair 1st and 4th
There are 12 socks that he can pick 1st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There are 8 socks that he can pick 3rd (all but pair of 1st and pair of
2nd).
There is 1 sock that he can pick 4th (pair of 1st).
In this case, there are
Case 4: Matching pair 2nd and 3rd
There are 12 socks that he can pick 1st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There is 1 sock that he can pick 3rd (pair of 2nd).
There are 8 socks that he can pick 4th (all but pair of 1st).
In this case, there are
Case 5: Matching pair 2nd and 4th
There are 12 socks that he can pick 1st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There are 8 socks that he can pick 3rd (all but pair of 1st and pair of
2nd).
There is 1 sock that he can pick 4th (pair of 2nd).
In this case, there are
Case 6: Matching pair 3rd and 4th
There are 12 socks that he can pick 1st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There are 8 socks that he can pick 3rd (all but pair of 1st and pair of
2nd).
There is 1 sock that he can pick 4th (pair of 3rd).
In this case, there are
In each of the six cases, there are
Therefore, the overall probability is
Solution 2
Since there are 12 socks in the drawer, there are
Next, we count the number of ways in which the 4 chosen socks can
include exactly 1 matching pair.
There are 6 possible matching pairs that can be chosen.
The remaining 2 socks must come from 2 different pairs of the remaining
5 pairs.
If we label the remaining pairs A, B, C, D, E, there are 10 possible
two-pair combinations from which these 2 socks can be chosen: AB, AC,
AD, AE, BC, BD, BE, CD, CE, DE. (We could also use a binomial
coefficient to calculate this total.)
For each of the 2 pairs from which these 2 socks will be chosen, there
are in fact 2 socks to choose.
Thus, the total number of ways in which the 4 socks can be chosen that
satisfy the given conditions is
This means that the probability is
Solution 3
Since there are 12 socks in the drawer, there are
These 4 socks can include 0 matching pairs, 1 matching pair, or 2
matching pairs.
We count the number of ways in which the 4 chosen socks can include 0
matching pairs or 2 matching pairs and subtract this from the total
number of ways of choosing 4 socks to obtain the number of ways of
choosing 1 matching pair.
To choose 0 matching pairs, we choose 4 of the 6 pairs and choose 1
of the 2 socks in each pair.
There are
Thus, there are
To choose 2 matching pairs, we choose 2 of the 6 pairs and choose
both socks in each pair.
There are
Thus, there are
This means that there are
This means that the probability is
Answer:
Since Karol uses 28 cups of Gluze, and the recipe calls for 4
cups of Gluze, he must have repeated the recipe
Since the recipe calls for 3 cups of Blurpos, and the recipe was
repeated 7 times, he must have used
Suppose that Karol uses 48 cups of Gluze.
Since
Suppose instead that Karol uses 48 cups of Blurpos.
Since
Therefore, the two possible values of
Karol starts with 64 cups of Gluze and 42 cups of Blurpos. One of
these ingredients will be the “limiting ingredient”; that is, when Karol
repeats his recipe, he will probably run out of one of the ingredients
before the other.
If Karol used all 64 cups of Gluze, then he must have used his recipe
If Karol used all 42 cups of Blurpos, then he must have used his recipe
Therefore, he can make the recipe at most 14 times since after the 14th
time, he would run out of Blurpos.
Each time Karol uses his recipe, he makes 60 Zippies.
When he uses his recipe 14 times, he makes
We start by assuming that Karol makes 1 recipe worth of
Zippies.
In other words, suppose that Karol makes 60 Zippies using 4 cups of
Gluze and 3 cups of Blurpos.
He sells each Zippie for $0.50, which earns him
His profit on each Zippie is $0.30, which means that his total profit is
Since he earns $30.00 and his profit is $18.00, his costs must be
Since Karol pays $1.80 for each cup of Gluze, he pays a total of
This means that he pays
Since he uses 3 cups of Blurpos, Karol pays
Since
Thus,
Now, the area of trapezoid
Since
The area of trapezoid
(We could instead note that trapezoid
Therefore,
Since
Finally, by the Pythagorean Theorem,
Draw a perpendicular from
Quadrilateral
Therefore,
Since
Therefore,
We note that
By the Pythagorean Theorem in
Finally, the area of trapezoid
Suppose that the centre of the circle of radius
Join
Join
As in (b),
Since
Since
By the Pythagorean Theorem in
Thus,
Using a similar analysis,
Thus,
From (b),
Therefore,
Finally,
The key move in Beryl’s strategy is to choose 2 on turn #2 (her
first turn).
On turn #1, Alphonse must choose 1, for a running total of 1.
On turn #2, Beryl chooses 2, for a running total of 3.
On turn #3, Alphonse can choose 1, 2 or 3, for a running total of 4, 5
or 6, respectively.
On turn #4, Beryl can choose 1, 2, 3, or 4, and so chooses 4, 3 or 2
(corresponding to Alphonse’s choices of 1, 2 or 3, respectively) to
achieve a running total of 8 in each case.
Therefore, if Beryl chooses 2 on turn #2 and then chooses appropriately
on turn #4, she is guaranteed to win.
Alphonse has a winning strategy when
To show this, we use the fact that
On turn #1, Alphonse must choose 1.
On turn #2, Beryl can choose 1 or 2; on turn #3, Alphonse can choose
1, 2 or 3.
No matter what Beryl chooses on turn #2, Alphonse can choose a number to
make the sum of their numbers on those two turns equal to 4.
(If Beryl chooses 1, Alphonse chooses 3. If Beryl chooses 2, Alphonse
chooses 2.)
Thus, Alphonse can ensure that the running total after 3 turns is
On turn #4, Beryl can choose 1, 2, 3, or 4; on turn #5, Alphonse can
choose 1, 2, 3, 4, or 5.
No matter what Beryl chooses on turn #4, Alphonse can choose a number to
make the sum of their numbers on those two turns equal to 5.
(If Beryl chooses 1, 2, 3, or 4, Alphonse chooses 4, 3, 2, or 1,
respectively.)
Thus, Alphonse can ensure that the running total after 5 turns is
On turn #6, Beryl can choose 1, 2, 3, 4, 5, or 6; on turn #7,
Alphonse can choose 1, 2, 3, 4, 5, 6, or 7.
No matter what Beryl chooses on turn #6, Alphonse can choose a number to
make the sum of their numbers on those two turns equal to 7.
(If Beryl chooses
Thus, Alphonse can ensure that the running total after 7 turns is
Note as well that, following this method, the largest that Beryl can
make the running total after turn #6 is
Therefore, Alphonse has a winning strategy when
When
When
When
When
When
We saw what happens when
We can continue these types of argument to show that Alphonse has a
winning strategy when
So far, we have:
Winning strategy for Alphonse | Winning strategy for Beryl |
---|---|
These small values of
We note that
Based on our hypotheses, our guess is that Beryl has a winning strategy
for
In order to prove that this is true, we need to show that none of
Consider
Note that
Turn(s) | Details | Condition |
---|---|---|
#1 | A must choose 1 | |
#2 and #3 | B can choose 1 or 2 A chooses 2 or 1, respectively |
Sum over two turns is 3 |
#4 and #5 | B can choose 1, 2, 3, or 4 A chooses 4, 3, 2, or 1, respectively |
Sum over two turns is 5 |
# ( |
B can choose A chooses |
Sum over two turns is |
Therefore, after 89 turns, using this strategy, Alphonse can ensure
that the running total is
Consider the case where
Let
Alphonse’s winning strategy for these values of
By doing this, after turn #89, Alphonse can ensure that the running
total is
(Note that in the
Consider
Note that
Turn(s) | Details | Condition |
---|---|---|
#1 and #2 | A must choose 1 B chooses 1 |
Sum over two turns is 2 |
#3 and #4 | A can choose 1, 2 or 3 B chooses 3, 2 or 1, respectively |
Sum over two turns is 4 |
#5 and #6 | A can choose 1, 2, 3, 4, or 5 B chooses 5, 4, 3, 2, or 1, respectively |
Sum over two turns is 6 |
# ( |
A can choose B chooses |
Sum over two turns is |
Therefore, after 90 turns, using this strategy, Beryl can ensure that
the running total is
Consider
Note that
By following the idea of the cases above, in particular making the sum
of the numbers on turn #1 and turn #2 equal to 2, the sum of the numbers
on turn #3 and turn #4 equal to 4 or 5, and the sum of the numbers on
the pairs of turns #5 and #6 through #87 and #88 equal to
In summary, we have shown that Beryl has a winning strategy for