Thursday, April 13, 2016
(in North America and South America)
Friday, April 14, 2016
(outside of North American and South America)
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The first bucket contains 7 red discs.
Each bucket after the first contains 3 more red discs than the previous bucket.
Thus, the second bucket contains
Solution 1
Let
Since the first bucket contains 17 green discs and each bucket after the first contains 1 more green disc than the previous bucket, then there are
Since the first bucket contains 7 red discs and each bucket after the first contains 3 more red discs than the previous bucket, then there are
The number of green discs in a bucket is equal to the number of red discs inside the same bucket when
Thus, there are an equal number of red discs and green discs in the bucket 5 after the first bucket, which is the 6
Solution 2
Using the fact that the first bucket contains 17 green discs and 7 red discs, and each bucket after the first contains 1 more green disc and 3 more red discs than the previous bucket, then we may summarize the number of green and red discs inside each bucket.
Bucket Number | Number of green discs | Number of red discs |
---|---|---|
1 | 17 | 7 |
2 | 18 | 10 |
3 | 19 | 13 |
4 | 20 | 16 |
5 | 21 | 19 |
6 | 22 | 22 |
Therefore, the 6
(Note that since the number of red discs is increasing by 3 each bucket and the number of green discs is increasing by 1 each bucket, then this is the only bucket in which the number of red discs and green discs will be equal.)
Solution 3
In the first bucket, there are 17 green discs and 7 red discs. Each bucket after the first contains 1 more green disc and 3 more red discs than the previous bucket.
Since there are 2 more red discs than green being put in, then the difference between the numbers of green and red discs will decrease by 2 for each bucket after the first.
Since the original difference is
Therefore, the 6
As in part (b), there are
The number of red discs in a bucket is equal to twice the number of green discs inside the same bucket when
In the 27
The total number of discs in this bucket is
A plate with 36 shaded squares has 10 shaded squares along each side of the plate, as shown.
We can see this from the diagram, or by considering a plate with
In this case, we can count
This means that there are
Here, we want
There are 10 squares along each side of the plate and the side length of the square plate is 60 cm, thus the side length of each of the shaded squares is
Since the plate is a square, and there are an equal number of identical shaded squares along each edge of the plate, then the unshaded area in the centre of the plate is also a square.
The area of this unshaded square in the centre of the plate is 1600 cm
Consider the row of squares along the left edge of the plate.
Since the side length of the square plate is 60 cm and the side length of the inner square is 40 cm, then the sum of the side lengths of the two shaded corner squares is
Therefore, each shaded corner square (and thus each shaded square) has side length
Using the same argument as in part (b), the area of the unshaded square in the centre of the plate is 2500 cm
The side length of the square plate is 60 cm and so the sum of the side lengths of 4 shaded squares (2 stacked vertically in the top two rows and 2 stacked vertically in the bottom two rows) is
Therefore, each of these shaded squares (and thus each shaded square on the plate) has side length
The side length of the square plate is 60 cm and each shaded square has length
There are 2 rows that each contain 24 shaded squares along each of the top and bottom of the square, and 2 additional rows that each contain
That is, the total number of shaded squares on the plate is
Triangle
Since
In
Therefore,
The shaded region lies inside the circle and outside the hexagon and thus its area is determined by subtracting the area of the hexagon from the area of the circle.
First we find the area of the hexagon.
Each vertex of hexagon
Since the circle has centre
Each side length of the hexagon is also 6, and so the hexagon is formed by six congruent equilateral triangles with side length 6. (For example,
Each of these triangles is congruent to
The area of each of the six congruent triangles is
Therefore, the area of hexagon
The area of the circle with centre
Finally, the area of the shaded region is
Let the area of the shaded region that we are required to find be
Let the area of the shaded region in the diagram to the right be
We may determine
First we determine
Consider the circle with centre
That is,
In
Since
In
Since
Therefore, the area of
Next, we determine the area of sector
Since
Thus, the area of sector
Therefore,
Finally, one-half of the area of the circle with centre
Simplifying further, the exact area of the shaded region is
The prime factorization of
Given an input of
Given an input of
We are told that this output is equal to 135.
Since
Since
If
However
If
Therefore, the only pair
Solution 1
Given an input of
Comparing these gives
Since
Further,
Therefore,
Similarly,
Therefore,
Finally,
Therefore,
Since
That is, each of the terms
Therefore,
If
If
If
If
Therefore, the triples
Solution 2
Given an input of
Comparing these gives
Since
Since
Since
Therefore, the equation
Since
Since
We proceed in a number of steps.
Step 1: Each exponent in the prime factorization must be a multiple of its prime
We show that, since the output is an integer multiple of the input, each exponent in the prime factorization must be an integer multiple of its corresponding prime.
This generalizes what we found in Solution 2 to part (c).
In other words, suppose that
By definition, the output of the Barbeau Process is
Note that we are allowing each of
Here, we are told that the output of the Barbeau process is
Therefore,
Rearranging, we obtain
Multiplying both sides by
Since the prime numbers
A similar argument will show that
Step 2:
Suppose that
In other words, suppose that
In this case, the factor
In this case,
But
Therefore,
Step 3: Simplifying the algebra
Combining Steps 1 and 2,
As in Solution 2 to part (c), we set
Therefore, the equation
Therefore, we want to find all of the non-negative integer solutions to
Step 4: Restrictions on
Since
If
Therefore,
If
Therefore,
Step 5: Determining the values of
Since
(If one value is 3, the rest must be 0. If one value is 2, then there must be one 1 and two 0s. If no value is 2, then there must be three 1s and one 0.)
Using the facts that
Category | Less than |
|||||
---|---|---|---|---|---|---|
(i) | 3 | 0 | 0 | 0 | Yes | |
(i) | 0 | 3 | 0 | 0 | Yes | |
(ii) | 2 | 1 | 0 | 0 | Yes | |
(ii) | 2 | 0 | 1 | 0 | Yes | |
(ii) | 2 | 0 | 0 | 1 | Yes | |
(ii) | 1 | 2 | 0 | 0 | Yes | |
(ii) | 0 | 2 | 1 | 0 | Yes | |
(ii) | 0 | 2 | 0 | 1 | Yes | |
(ii) | 1 | 0 | 2 | 0 | Yes | |
(ii) | 0 | 1 | 2 | 0 | Yes | |
(ii) | 0 | 0 | 2 | 1 | No | |
(iii) | 1 | 1 | 1 | 0 | Yes | |
(iii) | 1 | 1 | 0 | 1 | Yes | |
(iii) | 1 | 0 | 1 | 1 | No | |
(iii) | 0 | 1 | 1 | 1 | No |
In each case, we can check whether the possible value of
In summary, the possible values of