POTW #30 Challenge #1 Solution:
POTW #30 Challenge #2 Solution:
POTW #31 Question Gr. 8:
POTW #31 Question Gr. 7:
Thursday, May 2, 2019
POTW #31 - Number Sense Practice
The current problems (#31) help to support challenges in our current number sense (Integers) unit. And remember Gauss contest is May 15th. Examples and past tests are online and I (Mr. Milette) have hard copies of previous years' tests if you'd like to borrow. Just see me!
POTW #30 Challenge #1 Solution:
POTW #30 Challenge #2 Solution:
POTW #31 Question Gr. 8:
POTW #31 Question Gr. 7:
POTW #30 Challenge #1 Solution:
POTW #30 Challenge #2 Solution:
POTW #31 Question Gr. 8:
POTW #31 Question Gr. 7:
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The Other Grade 7/8 POTW (from last week that I did not do thet I will do now)
ReplyDeleteFirst, we need to find the value of b. To do this easily, we can convert the maximum (the tip of the parabola) into an equation. We can use vertex form with the variables h and k, where h is number of units the maximum moved to the right (from the origin) and k is the number of units the maximum moved up or down (from the origin). Since neither the numbers in the maximum’s coordinates are negative, we can carry on.
The vertex form states that = a(x-h)^2 + k, where a is the value of stretch/compression (dilation) from an original classic parabola (hard to explain what this is lol) to the current parabola. We are given h and k, so we can place it into the equation.
Y = a(x-7)^2 + 9 (notice that the 7 is now a negative. this is because the minus sign does not change. it would be a positive if the original 7 were negative)
Now that we have this base equation, we can substitute another point into the equation. We can only use point A because point B is not determined at this point in time.
8 = a(9-7)^2 + 9
8 = a(2)^2 + 9
8= 4a + 9
-1 = 4a
a = -(1/4)
Y = -(1/4) * (x-7)^2 +9 (wow that looks confusing online)
Substitute point B:
5 = -(1/4)*(b-7)^2 + 9 (subtract the 9)
-4 = -(1/4)*(b-7)^2 (divide both sides by -1/4)
16 = (b-7)^2 (square root both sides)
+/- 4 = b - 7
b= 11 or b = 3 (ok this is weird)
But since b must be to the left of 7, we know that b much be 3. Point B is (3,5). But while we’re still on this topic, the 11 would represent the coordinate directly across from point B, where the height is still 5. If we draw a line parallel to CD starting from point B, we get the coordinate (11,5). Anyways, this has nothing to do with the actual question, I used wanted to explain that.
Anyways, we know that the shape of shape ABCD (hehe) must have all striaght lines because it does not say otherwise. So we can drop the point A all the way down towards point (9,0) where the x-value stays the same and the y value changes.
This is point… F. Then we can drop down from point B to (3,0) to Point G… We can draw a line parallel to CD starting from Point B towards the line of AF. Point H at (9,5). Did this by taking A’s x value and B’s y value. Now we can add up the areas of triangle BCG, triangle ADF, triangle ABH and square BHGF.
Area = triangle ABH + triangle BCG + square BHGF + triangle AFD
Area = (3*6/2) + (2*5/2) + (5*6) + (4*8/2)
Area = 9 + 5 + 30 + 16
Area = 60
The area is 60 units ^2
Grade 8 POTW:
ReplyDelete(gather ‘round the campfire children, grandma’s telling a story)
one day i was bored on the morning bus so i decided to try and memorize some of the powers of 2 (meaning 2 is the base number and the exponent changes). but one thing i noticed is that 1) i haven’t been capitalizing my words and 2) every time the exponent is even, its a perfect square.
2^2 = 4 (perfect square of 2)
2^4 = 16 (perfect square of 4)
2^6 = 64 (perfect square of 8)
5^4 = (perfect square of 25)
Anyways, I thought I would use that knowledge in this POTW. First, we have to prime factorize the number 2020 (base number). 2020 = 2^2 x 5 x 101.
Now we can reword our equation to (2^2 x 5 x 101) ^2020. Since we have numbers in the bracket and an overall exponent, we can apply the exponent to each individual number.
(2^2)^2020) x (5^2020) x (101^2020).
(2^4040) x (5^2020) x (101^2020)
In these three brackets are 3 variables (a, b, c) which represent the number of even exponents in each bracket.
a = 2^4040
4040/2 = 2020 (but this isn’t inclusive, the math version not the wow join our game of tag version, meaning that we did not account for the lowest AND highest possible number. add 1 to make it better)
a = 2021
b = 5^2020
2020/2 = 1010 (same thing as above, add 1)
b = 1011
c = 101^2020
2020/2 = 1010
c = 1011
Now we multiply the numbers together which will get us some really large number.
2021 x 1011 x 1011 = 2,065,706,541
Well there you have it folks. Over 2 million divisors. great stuff right there
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ReplyDeleteGrade 8 POTW
ReplyDeleteI will go straight into it. To start things off, any integer to the power of a positive even integer (and 0) is a perfect square. To solve this problem, let’s first find the prime factorization of 2020. This is 2 x 2 x 5 x 101 = 2^2 x 5 x 101. Therefore, 2020^2020 = (2^2 x 5 x 101)^2020 = (2^2)^2020 x 5^2020 x 101^2020 = 2^4040 x 5^2020 x 101^2020. Based on this, every divisor of 2020 includes 2 to the power of a number greater than or equal to 0, and less than or equal to 4040, 5 to the power of a number greater than or equal to 0, and less than or equal to 2020, and 101 to the power of a number greater than or equal to 0, and less than or equal to 2020 (note that 0 is included as anything to the power of 0 is 1 with a few exceptions). Going off of the fact from before (any integer to the power of a positive even integer (and 0) is a perfect square), we can find the number of even integers there are between each of these possible exponents:
Base 2: 4040/2 = 2020 + 1 (from the 0) = 2021
Base 5: 2020/2 = 1010 + 1 (from the 0) = 1011
Base 101: 2020/2 = 1010 + 1 (from the 0) = 1011
Now we just need to multiply these to find out the number of perfect square divisors:
2021 x 1011 x 1011 = 2065706541
Therefore, there are 2,065,706,541 (two billion sixty five million seven hundred six thousand five hundred forty one) perfect square divisors of 2020^2020.
Grade 7/8 Challenge POTW #2 (Last Week)
ReplyDeleteJust like the other one, I will go straight into this. The peak or vertex of the parabola is (7,9). Using vertex form or y = a(x - h)^2 + k, we can get this equation:
y = a(x - 7)^2 + 9
We still need another point to solve the equation, and as B is not yet complete, we can use A:
8 = a(9 - 7)^2 + 9
8 = a(2)^2 + 9
8 = 4a + 9
-1 = 4a
-1/4 = 4a/4
-1/4 = a
We now have the parabola’s equation. Using this and B’s y value, we can solve for B’s x or b value:
y = (-1/4)(x - 7)^2 + 9
5 = (-1/4)(b - 7)^2 + 9
5 - 9 = (-1/4)(b - 7)^2 + 9 - 9
-4 = (-1/4)(b - 7)^2
-4/(-1/4) = (-1/4)(b - 7)^2/(-1/4)
16 = (b - 7)^2
√16 = √(b - 7)^2
4 = b - 7 or -4 = b - 7
If b - 7 = 4:
b - 7 = 4
b - 7 + 7 = 4 + 7
b = 11
If b - 7 = -4
b - 7 = -4
b - 7 + 7 = -4 + 7
b = 3
So b can be either 3 or 11. We know that b is less than 7 as it is to the left of the vertex. Therefore, b = 3. The coordinate of B is (3, 5). We now need to find the coordinates of point C and D. We know that their y coordinate is 0 as they both are on the “ground”. This means we have to find the parabola’s x intercept to solve for both coordinates. When y = 0, x =:
y = (-1/4)(x - 7)^2 + 9
0 = (-1/4)(x - 7)^2 + 9
0 - 9 = (-1/4)(x - 7)^2 + 9 - 9
-9 = (-1/4)(x - 7)^2
-9/(-1/4) = (-1/4)(x - 7)^2/(-1/4)
36 = (x - 7)^2
√36 = √(x - 7)^2
6 = x - 7 or -6 = x - 7
If x - 7 = 6:
x - 7 = 6
x - 7 + 7 = 6 + 7
x = 13
If x - 7 = -6
x - 7 = -6
x - 7 + 7 = -6 + 7
x = 1
Therefore, the x intercepts of this parabola is (1,0) and (13,0). C is to the left and D is to the right so their coordinates are (1,0) and (13,0) respectively. So the coordinates of each point are:
A: (9,8)
B: (3,5)
C: (1,0)
D: (13,0)
E: (9,8)
There are many ways to calculate the area of quadrilateral ABCD, but I will just show one. Okay, so if we draw a line straight down and perpendicular to the x axis, we get the coordinate (3,0). We can do the same to point A and get (9,0). We will label each F and G respectively. Now we have 3 shapes, trapezoid ABFG, triangle BCF, and triangle ADG. We will now calculate the areas of each and add them up:
Trapezoid ABFG:
((AG + BF)/2)FG = ABFG Area
((AG + BF)/2)FG = ABFG Area
((8 + 5)/2)(9 - 3) = ABFG Area
(13/2)6 = ABFG Area
(6.5)6 = ABFG Area
39 units^2 = ABDG Area
Triangle BCF:
(CF)(BF)/2 = BCF Area
(3 - 1)(5)/2 = BCF Area
(2)(5)/2 = BCF Area
5 units^2 = BCF Area
Triangle ADG:
(DG)(AG)/2 = ADG Area
(13 - 9)(8)/2 = ADG Area
(4)(8)/2 = ADG Area
16 units^2 = ADG Area
Now we can add them all up to get the total area of quadrilateral ABCD:
ABFG + BCF + ADG = ABCD Area
39 + 5 + 16 = ABCD Area
60 units^2 = ABCD Area
Therefore, the area of quadrilateral ABCD is 60 units squared.
Grade 7 POTW
ReplyDeleteThis POTW is pretty simple and straightforward. Firstly, 2^6 = 64. If (x - 1)^(x + y) = 64, and x and y are integers, this means that the base must be an integer and so must the exponent. Therefore, we just need to find all the possible integer bases and exponents that equal 64:
2^6 = 4^3 = 8^2 = 64^1 = -2^6 = 8^2 = 64
For each exponent, we just need to find the x and y value that equals the value:
2^6:
(x - 1) = 2
(x + y) = 6
x = 3
y = 3
4^3:
(x - 1) = 4
(x + y) = 3
x = 5
y = -2
8^2:
(x - 1) = 8
(x + y) = 2
x = 9
y = -7
64^1:
(x - 1) = 64
(x + y) = 1
x = 65
y = -64
-2^6:
(x - 1) = -2
(x + y) = 6
x = -1
y = 7
-8^2:
(x - 1) = -8
(x + y) = 2
x = -7
y = 9
There are six ordered pairs that satisfy this equation including (3, 3), (5, −2), (9, −7), (65, −64), (−1, 7), and (−7, 9).