Tuesday, February 25, 2020
(in North America and South America)
Wednesday, February 26, 2020
(outside of North American and South America)
©2020 University of Waterloo
Simplifying,
Answer: A
When
Alternatively,
Answer: D
Since
Since
Since
Therefore, the coordinates of
Answer: B
If
Also,
Lastly,
Therefore,
Answer: B
Since
Now
This means that
Using the information that we know,
(Alternatively, we could have noted that
Answer: E
From the bar graph, Matilda saw 6 goldfinches, 9 sparrows, and 5 grackles.
In total, she saw
This means that the percentage of birds that were goldfinches is
Answer: C
Since the average of
In order for
Since
Answer: C
To determine 30% of Roman’s $200 prize, we calculate
After Roman gives $60 to Jackie, he has
He splits 15% of this between Dale and Natalia.
The total that he splits is
Since Roman splits $21 equally between Dale and Natalia, then Roman gives Dale a total of
Answer: A
The 1st row has 0 shaded squares and 1 unshaded square.
The 2nd row has 1 shaded square and 2 unshaded squares.
The 3rd row has 2 shaded squares and 3 unshaded squares.
The 4th row has 3 shaded squares and 4 unshaded squares.
Because each row has 2 more squares than the previous row and the squares in each row alternate betweeen unshaded and shaded, then each row has exactly 1 more shaded square than the previous row.
This means that, moving from the 4th row to the 2020th row, a total of
Thus, the 2020th row has
Answer: D
We extend
Because quadrilateral
Thus,
Since
Now
By the Pythagorean Theorem, since
Since
Finally, we can calculate the area of
The area of
The area of rectangle
Therefore, the area of pentagon
Answer: E
Since
Answer: A
Suppose that the length, width and height of the prism are the positive integers
Since the volume of the prism is 21, then
We note that each of
The positive divisors of 21 are 1, 3, 7, and 21, and the only way to write 21 as a product of three different integers is
Therefore, the length, width and height of the prism must be 1, 3, and 7, in some order.
The sum of these is
Answer: A
Since
Thus, if
Answer: B
Since
In particular, the greatest possible value of
For the product of three numbers to be positive, either all three numbers are positive (that is, none of the numbers is negative) or one number is positive and two numbers are negative. (If there were an odd number of negative factors, the product would be negative.)
If all three numbers are positive, the product is as large as possible when the three numbers are each as large as possible. In this case, the greatest possible value of
If one number is positive and two numbers are negative, their product is as large as possible if the positive number is as large as possible (7) and the product of the two negative numbers is as large as possible.
The product of the two negative numbers will be as large as possible when the negative numbers are each “as negative as possible” (that is, as far from 0 as possible). In this case, these numbers are thus
So the greatest possible value of
Combining the two cases, we see that the greatest possible value of
Answer: A
Since the ratio of green marbles to yellow marbles to red marbles is
Since 63 of the marbles in the bag are not red, then the sum of the number of green marbles and the number of yellow marbles in the bag is 63.
Thus,
Answer: B
Let
Let
Since the square has a right-angle at each of its vertices, then
By the Pythagorean Theorem,
In terms of
Since we are given that the area of the circle is
Since
In terms of
Answer: E
Suppose that Carley buys
In total, Carley will then have
Since the contents of each bag was the same and Carley made no incomplete treat bags and there were no left-over candies, then it must be the case that
We want to find the minimum possible positive value of
Dividing by the common factor of 5, the equation
Since
Since 10, 20 and 30 are not multiples of 8, and 40 is a multiple of 8, then the smallest possible value of
In this case,
Since
Thus, the minimum number of boxes that Carley could have bought is
Answer: B
Solution 1
Suppose that when Nate arrives on time, his drive takes
When Nate arrives 1 hour early, he arrives in
When Nate arrives 1 hour late, he arrives in
Since the distance that he drives is the same in either case and distance equals speed multiplied by time, then
Expanding, we obtain
The total distance that Nate drives is thus
When Nate drives this distance in 5 hours at a constant speed, he should drive at
Solution 2
Suppose that the distance that Nate drives is
Since driving at 40 km/h causes Nate to arrive 1 hour late and driving at 60 km/h causes Nate to arrive 1 hour early, then the difference between the lengths of time at these two speeds is 2 hours.
Since time equals distance divided by speed, then
Thus, the distance that Nate drives is 240 km.
At 40 km/h, the trip takes 6 hours and Nate arrives 1 hour late.
To arrive just in time, it should take 5 hours.
To drive 240 km in 5 hours, Nate should drive at a constant speed of
Answer: D
For each of the 10 questions, each correct answer is worth 5 points, each unanswered question is worth 1 point, and each incorrect answer is worth 0 points.
If 10 of 10 questions are answered correctly, the total score is
If 9 of 10 questions are answered correctly, either 0 or 1 questions can be unanswered. This means that the total score is either
If 8 of 10 questions are answered correctly, either 0 or 1 or 2 questions can be unanswered. This means that the total score is either
If 7 of 10 questions are answered correctly, either 0 or 1 or 2 or 3 questions can be unanswered. This means that the total score is one of 35, 36, 37, or 38 points.
If 6 of 10 questions are answered correctly, either 0 or 1 or 2 or 3 or 4 questions can be unanswered. This means that the total score is one of 30, 31, 32, 33, or 34 points.
So far, we have seen that the following point totals between 30 and 50, inclusive, are possible:
If 5 or fewer questions are answered correctly, is it possible to obtain a total of at least 39 points?
The answer is no, because in this case, the number of correct answers is at most 5 and the number of unanswered questions is at most 10 (these both can’t happen at the same time) which together would give at most
Therefore, there are exactly 6 integers between 30 and 50, inclusive, that are not possible total scores.
Answer: D
We determine when
To do this, we first look individually at the units digits of
The units digits of powers of 3 cycle
To see this, we note that the first few powers of 3 are
This means that the units digits of powers of 3 cycle every four powers of 3.
Therefore, of the 100 powers of 3 of the form
The units digits of powers of 7 cycle
To see this, we note that the first few powers of 7 are
Since
Therefore, of the 105 powers of 7 of the form
For
the units digit of
the units digit of
the units digit of
the units digit of
The number of possible pairs
Answer: E
To determine the number of points
In other words, we determine the number of integers
We note that, as
Also, when
Further, when
Therefore,
There are
Answer: D
Since
We can thus apply the Pythagorean Theorem to obtain
Since
Consider
Each is right-angled and they share a common angle at
This tells us that
Using the lengths that we know,
Finally,
Answer: B
Suppose that
The first digit of
Since there are three 1s in
The first digit of
Thus,
Case 1:
Since no two 2s can be next to each other, we next place the remaining two 2s.
Reading from the left, these 2s can go in positions 4 and 6, 4 and 7, 4 and 8, 4 and 9, 5 and 7, 5 and 8, 5 and 9, 6 and 8, 6 and 9, or 7 and 9.
In other words, there are 10 possible pairs of positions in which the 2s can be placed.
Once the 2s are placed, there are 5 positions left open.
Next, we place the remaining two 1s.
If we call these 5 positions A, B, C, D, E, we see that there are 10 pairs of positions in which the 1s can be placed: A and B, A and C, A and D, A and E, B and C, B and D, B and E, C and D, C and E, D and E.
This leaves 3 positions in which the two 3s and one 4 must be placed. Reading from the left, a 3 must be placed in the first empty position, since there must be a 3 before the 4.
This leaves 2 positions, in which the remaining digits (3 and 4) can be placed in any order; there are 2 such orders (3 and 4, or 4 and 3).
In this case, there are
Case 2:
Since no two 2s can be next to each other, we next place the remaining 2s.
Reading from the left, these 2s can go in positions 5 and 7, 5 and 8, 5 and 9, 6 and 8, 6 and 9, or 7 and 9.
In other words, there are 6 possible pairs of positions in which the 2s can be placed.
Once the 2s are placed, there are 4 positions left open.
Next, we place the remaining 1. There are 4 possible positions in which the 1 can be placed.
This leaves 3 positions in which the two 3s and one 4 must be placed. Reading from the left, a 3 must be placed in the first empty position, since there must be a 3 before the 4.
This leaves 2 positions, in which the remaining digits (3 and 4) can be placed in any order; there are 2 such orders.
In this case, there are
Case 3:
Since no two 2s can be next to each other, we next place the remaining 2s.
Reading from the left, these 2s can go in positions 6 and 8, 6 and 9, or 7 and 9.
In other words, there are 3 possible pairs of positions in which the 2s can be placed.
Once the 2s are placed, there are 3 positions left open with only the two 3s and the 4 remaining to be placed.
Reading from the left, a 3 must be placed in the first empty position, since there must be a 3 before the 4.
This leaves 2 positions, in which the remaining digits (3 and 4) can be placed in any order; there are 2 such orders.
In this case, there are
Combining the three cases, we see that there are
Answer: C
Suppose that
Since the edge length of the cube is 200, then
We consider tetrahedron (that is, triangle-based pyramid)
The volume of a tetrahedron is equal to one-third times the area of its triangular base times the length of its perpendicular height.
First, we consider tetrahedron
Thus, the volume of
Next, we consider tetrahedron
From the given information, the shortest distance from
We need to calcualte the area of
Since
This means that
Let
Then
Since
By the Pythagorean Theorem,
This means that the volume of tetrahedron
This means that
Of the given answers, this is closest to 59, which is (D).
Answer: D
We will use the result that if a positive integer
This result is based on the following facts:
F1. Every positive integer greater than 1 can be written as a product of prime numbers in a unique way. (If the positive integer is itself prime, this product consists of only the prime number.) This fact is called the “Fundamental Theorem of Arithmetic”. This fact is often seen implicitly in finding a “factor tree” for a given integer. For example,
F2. If
F3. If
From these facts, the positive divisors of
This means that there are
Similarly, there are
Since every combination of these possible values gives a different divisor
Suppose that
We have written
This means that
The positive divisors of 64 are 1, 2, 4, 8, 16, 32, 64.
Of these, only 1, 2, 4 are divisors of 60.
Therefore,
Since each of
Case 1:
The only ways in which 4 can be written as a product of positive integers each at least 2 are
Thus, either
Since
Substituting into the equation
Since
Therefore, we could have
Combining with the possible values of
We can verify that
Case 2:
The only way in which 2 can be written as a product of positive integers each at least 2 is
Thus,
Since
Alternatively, knowing that
32 | 1 | 30 | 0 | 31 | 2 | 62 |
16 | 2 | 14 | 1 | 15 | 3 | 45 |
8 | 4 | 6 | 3 | 7 | 5 | 35 |
4 | 8 | 2 | 7 | 3 | 9 | 27 |
2 | 16 | 0 | 15 | 1 | 17 | 17 |
1 | 32 | -1 | 31 | 0 | 33 | 0 |
There are no values of
Case 3:
Since each factor on the left side is supposedly at least 2, what can this mean? This actually means that there are no factors on the left side. In other words,
(See if you can follow the argument before Case 1 through to verify that there are no contradictions.)
Here,
64 | 1 | 62 | 0 | 63 | 2 | 126 |
32 | 2 | 30 | 1 | 31 | 3 | 93 |
16 | 4 | 14 | 3 | 15 | 5 | 45 |
8 | 8 | 6 | 7 | 7 | 9 | 63 |
4 | 16 | 2 | 13 | 3 | 15 | 51 |
2 | 32 | 0 | 31 | 1 | 33 | 33 |
1 | 64 | -1 | 63 | 0 | 65 | 0 |
There are no values of
Therefore, combining the results of the three cases, the positive integer
Since
It remains to determine the number of pairs of primes
In the first case, the possibilities are:
In the second case, the only possibility is
This means that there are 4 possible values of
Answer: A