Summer Problem Solving Marathon Question #47

[Value = 3 points]

A certain polynomial has a remainder of 3 when divided by x-1, and a remainder of 5 when divided by x-3. What is its remainder when divided by x² - 4x + 3?

Wrapping Up the Summer Problem Solving Marathon

We're going to end the Summer Problem Solving Marathon at the end of this week, which will take us through Problem #50. Math team meetings will resume on Tuesday September 6.

Summer Problem Solving Marathon Question #46

[Value = 3 points]

What is the surface area of a cube with a space diagonal of length 1? (The space diagonal of a cube is the line segment from a vertex of the cube to the most distant vertex from that vertex.)

Summer Problem Solving Marathon Question #45

(This question should have been posted Thursday. Sorry for the delay.)

[Value = 6 points]

A fair coin is tossed 12 times. What is the probability that there are no two consecutive heads in the string of tosses?

Summer Problem Solving Marathon Solution #43

Let log₈(log₂ x) = y. Then log₂ x = 8^y = 2^3y, and x = 2^23y.

We then also have log₂(log₈ x) = y, so log₈ = 2^y, and x = 8^2y = 2^3(2y).

Since log₈(log₂ x) = log₂(log₈ x), we have 2^23y = 2^3(2y). Thus 3^3y = 3(2^y), and 2^2y = 3, and 2y = log₂ 3, and y = (log₂3)/2.

Since x = 2^3(2y), log₂ x = 3(2^y), and (log₂ x)² = 9(2^2y).

We now substitute (log₂3)/2 for y in this expression. This gives 9(2^log₂3) = 9 x 3 = 27.

Summer Problem Solving Marathon Solution #42

Let x be the probability of getting heads (H) on the biased coin. Then 1-x is the probability of getting tails (T).

We first calculate the probability of getting exactly one head. There are five outcomes with exactly one head (matching the five ways of picking a location for the H in the sequence). Each of those five ways has probability x(1-x)⁴ of occurring. So p(H=1) = 5x(1-x)⁴.

Next we calculate the probability of getting exactly two heads. There are ten outcomes with exactly two heads (matching the ten ways of picking two locations for the two Hs in the sequence). Each of these ten ways has probability x²(1-x)³ of occurring. So p(H=2)=10x²(1-x)³.

Since p(H=1) = p(H=2), we have:

5x(1-x)⁴ = 10x²(1-x)³

Simplifying (given that 0 < x < 1), we have:

1 - x = 2x
x = 1/3

We now calculate p(H=3). There are ten ways of picking locations for the three heads. Each of those ten ways has probability x³(1-x)² of occurring. So our probability is:

10(1/3)³(2/3)² = 40/243

Summer Problem Solving Marathon Question #44

[Value = 5 points]

Simplify (56 + 6√43)^3/2 - (56 - 6√43)^3/2 to an integer value. No calculators allowed for this question!