Welcome to Math Team!
2³/(7x2)/log₃3¹⁷ = 8/14/17 = .0336134454
WHO ARE WE?
WHAT DO WE DO?
ANNOUNCEMENTS
Problem Number FUN
Does this look familiar:
72
27
45
39
36
??
21
18
99
13
28
21
7
Solution
The answer is actually 12. To find the answer, all you have to do is add the digits!
Problem Number 1
Solution
Answer: 32 has 6 factors; 50 also has 6 factors; 255 has 8 factors
Here is a trick for learning to count factors:
Write the number in a prime factorization format. For example, 32 is 2⁵. Then to find the number of factors you take the exponents of the prime factors and add 1, then multiply (32 only has one prime factor so you don’t need to multiply).
Try 50 and 255 using this method! See if you get 6 and 8 respectively.
Problem 2: Angry Kwam
Jayson likes to play chess. When he was playing one day against Kwam, Jayson checkmated him, causing Kwam to bite two opposite corners off of the chessboard in a furious rage. Jayson, seeing the useless checkerboard, decides to use it to store his dominos. How many ways can Jayson completely cover up his broken chessboard with dominoes? Each domino takes an area of 2x1 squares, and two ways are different if they cannot be rotated to become equal.
Solution
Answer: No Solution!; 0 ways
Each domino takes up 2 1x1 squares, one white square and one black square. If you remove two opposite corners, those are the same color, so you are left with 30 black and 32 white squares (or vice versa). 32 does not equal 30.
Problem Number 3
What is 1/2 + 1/4 + 1/8 +1/16 +...?
What is 3/8 + 6/16 + 9/32 + 12/64 + 15/128+....?
Solution to the First Series:
Answer= 1
For the first geometric series can be added using the formula: a/(1-r), where a is the first term and r is the common ratio.
Plug in ½ for a and ½ for r to get 1.
Now for the fun one!
Solution to the Second Series:
What is 3/8 + 6/16 + 9/32 + 12/64 + 15/128+....?
Notice that the series above is ALMOST an infinite geometric series, but the numerator keeps changing. So how do we fix that?
Set 3/8 + 6/16 + 9/32 + 12/64 + 15/128+... = S
Take that S and multiply by ½
½ S= 3/16 + 6/32 + 9/64 + 12/128+...
Notice anything?
Solution to the Second Series:
What is 3/8 + 6/16 + 9/32 + 12/64 + 15/128+....?
3/8 + 6/16 + 9/32 + 12/64 + 15/128+... = S
3/16 + 6/32 + 9/64 + 12/128+... = ½ S
Lets subtract the two series:
3/8 +3/16 +3/32 +3/64+ 3/128= ½ S
From here, just apply our trusty formula a/(1-r), where a is ⅜ and r is ½.
We find that ½ S is 3/4 .
Then S= 3/2.
Problem 4: 43, 37, 51, 89, 34, 10
Given the sequence 43, 37, 51, 89, 34, What is the next number?
Problem 4 Solution
In the title, it says the sequence is 43, 37, 51, 89, 34, 10.
ALWAYS READ THE ENTIRE PROBLEM!!!!
Problem not 86: Counting is Hard Sometimes
How many multiples of 3 are there between 53 and 313?
Problem 5 Solution
Answer: 87
The number of multiples of 3 between 53 and 313 is the same as the number of multiples of 3 between 54 and 312, inclusive. This is also the same as the number of multiples of 3 between 54 - 54 = 0 and 312 - 54 = 258. Out of the numbers from 0 to 258, we want to count 0, 3, 6, …, 255, 258. If we divide all of these numbers by 3, we get 0, 1, 2, …, 85, 86. There are 86 - 0 + 1 = 87 numbers from 0 to 86.
Problem 6: Drawing is Hard Sometimes Too
Jayson is attempting to draw a triangle with vertices A, B, and C. He wants to make side AB = 1 cm , side BC = 41 cm, and side CA = 42 cm. Assuming that Jayson draws at a rate of 1 cm per second, doesn’t take any breaks, and never gives up, how many seconds will it take him to draw a triangle of those side lengths?
Problem 6 Solution
Answer: ∞ seconds
Jayson’s just too bad at drawing. Well, in his defense, I suppose there’s probably not anyone in the world who’s good enough to draw that triangle. To see why, first draw CA (42 cm). If point B is not on line CA, then AB + BC must be greater than 42 (going in a straight line from A to C is obviously shorter than making a detour and going from A to B and then to C). Then, since AB + BC = 1 + 41 = 42, point B must be on line CA, making triangle ABC not a triangle and just a line. This is also called the triangle inequality which states that if the sides of a triangle are a,b, and c, then a + b > c.
Problem 11011-2
BASES: We operate in base 10 because we count from 0-9 and then go to the tens digit. Our system operates like this _ _ _ _ . _ _
In Base 10, the place values work like 103 102 101 100 10-1 10-2
(think about the expansion form of a number; you all probably learned it in middle school or elementary school)
Note that you can only use the digits 0-3 in base 4 _ _ _ _ . _ _
In Base 4, the place values would work like 43 42 41 40 4-1 4-2
Knowing this, what is 1004 written in base 10?
What about 100 written in base 4?
How about 200 written in base -2?
Solution
Answer: 16; 12124; and finally 111011000-2
For 1004 in base ten, simply use the place values to determine that there is 1 (42) and 0 of everything else. Then you know the value is 16 in base ten.
For 100 written in base 4, try and find the largest exponent of 4 that will fit in 100 (43= 64, anything larger would not fit) Then, continue down the exponents of 4, subtracting the largest amount that you can. 1(64) + 2(16) + 1(4) + 2 =100.
For 200 in base -2, this one will require a little bit of trial and error, but don’t be afraid to go over 200 and then subtract from that value.
Problem 8: Angry Kwam Part 2
Kwam is stuck in the center cube of a 3x3 rubik’s cube. He can only move to adjacent cubes, and has to find his way to any one of the 6 cubes adjacent to the center cube while passing through all the other cubes exactly once. How many ways can he do this? Two paths are different if they cannot be rotated to be the same path.
Problem 8 Solution
Answer: No solution; 0 ways
If you color code the cubes black and white alternating, there are 14 black and 13 white cubes (or vice versa).With 14 black cubes, the center cube is black, and the 6 cubes adjacent to it are white. In 26 moves, Kwam can only end up on a piece of the same color, but he is supposed to start on a black cube and end on a white cube.