BMAD

When factoring quadratics the most popular method we teach in high school and college is known as the AC method. The method is straightforward, given a quadratic ax^2 + bx + c, multiply a and c, find factors of AC that can add and make b, replace bx with those two factors written as additve statements, then group, factor, and combine). For example: 9x^2 -30x + 25 9 * 25 = 225 15 * 15 = 225 and 15 + 15 = 30 9x^2 -15 x -15 x + 25 (9x^2 -15x) + (-15x + 25) ** 3x(3x-5) -5(3x-5) (3x-5)(3x-5) Now, this approach works for many people but there are a large number of people who get stuck at the stage where I indicated the "**". Especially when it comes to factoring with the negative sign. So there is an alternative approach, I would like to present, that removes a lot of issues with factoring: 9x^2 -30x + 25 9 * 25 = 225 15 * 15 = 225 and -15 + -15 = -30 9x^2 / -15x and 9x^2 / -15x write each factor with x under ax^2 Reduce= 3x/-5 (Numerator + denominator) (3x-5)(3x-5) This approach seems much simpler and easier for students to handle and works every time when a,b, and c are mutually prime. Your task, prove why the second method works just as often as the first.

is this the best solution?

So with a more expensive pizza, Charlie pays less even though the value of each slice increases?

Alice, Bob, and Charlie order a pizza for $10. It turned out that Alice and Bob were very hungry, eating 6 slices a piece, and Charlie just had 2 slices. They want to split the pizza cost fairly, but they are subject to a practical issue. No one has coins to make change, so they have to round how much each owes to the nearest dollar. What’s the fair way to split the bill? What if the pizza was $11?

fair point. I was thinking twins would be the exception but I agree with you Phil.

But you are right minus a trivial case.

Is there no cases where this isn't true?

Is it true that no matter how much older one sibling is to the other; eventually, the younger sibling will be half as old as the older?

Suppose you pick two random numbers less than n, then [n/2]2 pairs are both divisible by 2. [n/3]2 pairs are both divisible by 3. [n/5]2 pairs are both divisible by 5. ... (Here [x] is the greatest integer less than or equal to x, usually called the floor function.) So the number of relatively prime pairs less than or equal to n is (by the inclusion/exclusion principle): n 2 - sum([n/p]2) + sum([n/pq]2) - sum([n/pqr]2) + ... here the sums are taken over the primes p,q,r,... less than n. Letting mu(x) be the möbius function this is so the desired constant is the limit as n goes to infinity of this sum divided by n2, or sum(mu(k)/k2) (sum over positive integers k). But this series times the sum of the reciprocals of the squares is one, so the sum of this series, the desired limit, is 6/2. This number is approximately sum(mu(k)[n/k]2) (sum over positive integers k) 60.7927...%

It seems that I disagree as there should always be a way for the surgeon to win.

yeah I believe it is actually 6/pi2.

lol

Let E be the elliptic curve defined by y2 = x3 + x + 1 in Z11xZ11: Find the points in E. Compute 2(3,3) and -2(3,3)+(2,0) Suppose in the ElGamal using the curve in 1, Alice constructs P= (8,9) and Q=2P=(0,1) and makes public P,Q and E. She receives S= (3,3), T=(2,0) from Bob. Recover the message m.

A random walk on the three dimensional integer lattice is defined as follows. The walker starts at (0, 0, 0). A standard six sided die is rolled six times. After each roll the walker moves to one of its six nearest neighbors, according to the following protocol: if the die rolls 1, 2, 3, 4, 5, or 6 dots the walker jumps one unit in the +x, −x, +y, −y, z, −z direction respectively. Find the probability that after the sixth roll the walker is back at its starting point (0, 0, 0).

There are M gold fish and K silver fish in a lake. They are caught and eaten one at a time at random until only one color of fish remains in the lake. One of the silver fish is named George. Find the probability George is not eaten.

Hmmm 1/19 is what I found.

I have a totally different answer. So I assume, (a).

Determine the largest number d such that the following is true: If the points of the perimeter of an equilateral triangle of side 1 are colored with four colors, then there must be two points of the same color which are at least distance d apart.

Form a “triangle” with 10 blocks in its top row, 9 blocks in the next row, etc., until the bottom row has one block. Each row is centered below the row above it. Color the blocks in the top row red, white and blue in any way. Use these two rules to color the remaining rows of the triangle: • If two consecutive blocks in a row have the same color, the block between them in the row below has the same color. • If two consecutive blocks in a row have different colors,the block between them in the row below has the third color. Tell how you can always predict the color of the bottom block after seeing only the top row (and not constructing the intermediate rows).

AHH.. I see the error in my op. What I mean is: Show that 3w - 1 is the GCD of 3m - 1 and 3n - 1; i mistakenly raised the ' - 1 '