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Let's start by finding the prime numbers greater than 10 that Syd could have chosen. The prime numbers between 11 and 100 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.

Since Syd is choosing two different primes, he has 20 options for the first prime and 19 options for the second. The resulting products would be:

11 * 13 = 143

11 * 17 = 187

11 * 19 = 209

...

89 * 97 = 8613

So there are 20 * 19 = 380 different products that Syd could have ended up with.

There are 24 * 23 = 552 possible pairs of tiles that Steve and Ellen can choose.

Out of these, there are 4 * 4 = 16 tiles that are both red.

So, the probability that both tiles are red is 16/552 = 4/138.

The problem can be restated as finding the smallest positive integer n such that there exist 100 distinct subsets of the club, each of size k, where 1 <= k <= n. This is equivalent to finding the smallest positive integer n such that n choose k is greater than or equal to 100 for all values of k from 1 to n.

By the Pigeonhole Principle, if we have n people, then at most n choose k people can review the same book. Hence, we want the smallest positive integer n such that:

C(n,1) >= 100

C(n,2) >= 100

...

C(n,n) >= 100

We can start by computing some of the binomial coefficients and finding the smallest value of n that satisfies the conditions.

C(2,1) = 2 >= 100 (not satisfied)

C(3,1) = 3 >= 100 (not satisfied)

C(3,2) = 3 >= 100 (not satisfied)

C(4,1) = 4 >= 100 (not satisfied)

C(4,2) = 6 >= 100 (not satisfied)

C(4,3) = 4 >= 100 (not satisfied)

C(5,1) = 5 >= 100 (not satisfied)

C(5,2) = 10 >= 100 (not satisfied)

C(5,3) = 10 >= 100 (not satisfied)

C(5,4) = 5 >= 100 (not satisfied)

The smallest n that satisfies all the conditions is 6, so the smallest positive integer n that allows us to avoid having the exact same group of members over all 100 days is 6.

The number of ways the baking club can form an executive committee with people is given by the binomial coefficient\( C(n,k)\), where n is the total number of people in the baking club (including Mark) and k is the number of people on the executive committee.

So, the number of ways the baking club can form an executive committee with people is \(C(n,k) = n! / (k! (n-k)!)\).

To solve this problem, we can use similar triangles. Let the side length of the cube be s, and let the midpoints of AB, AC, and BC be P, Q, and R, respectively. Then OP, OQ, and OR are the diagonals of the faces of the cube, and so they are equal to \($\sqrt{2}s$\).

Consider triangle AOB. By the Pythagorean theorem, we have \($OA^2 + OB^2 = 1 + 1 = 2$\), so \($AO^2 = 2$\). Since angle AOB is 90 degrees, we can use similarity to see that triangle OAP is similar to triangle OAB, so \($\frac{OA}{OP} = \frac{OB}{OP}$\). Solving, we find that \($OP = \frac{\sqrt{2}}{2}$\).

Using similar triangles, we can find that \($OQ = \frac{\sqrt{2}}{2}$ and $OR = \frac{\sqrt{2}}{2}$\). Since the sum of the squares of the lengths of the diagonals of a cube is equal to three times the square of the side length, we have

\($$(OP^2 + OQ^2 + OR^2) = 3s^2$$\)

Substituting the values we found for $OP$, $OQ$, and $OR$, we have

\($$\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = 3s^2$$\)

Simplifying and solving for \($s$\), we find that \($s = \frac{1}{3}$\). Thus, the side length of the cube inscribed in the tetrahedron is \($\frac{1}{3}$\).

For part (a), we can use the AM-GM (Arithmetic Mean-Geometric Mean) inequality, which states that for any positive real numbers a_1, a_2, ..., a_n, their arithmetic mean is always greater than or equal to their geometric mean:

\((a_1 + a_2 + ... + a_n) / n >= sqrt[(a_1 * a_2 * ... * a_n)^(1/n)]\)

Using this inequality repeatedly, we can prove the given inequality as follows:

\((a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) = [(a^2 + b^2)^2][(c^4 + d^4)(e^4 + f^4)]\)

By AM-GM,

\([(a^2 + b^2)^2][(c^4 + d^4)(e^4 + f^4)] >= [(a^2 + b^2)^2][sqrt[(c^4 * d^4 * e^4 * f^4)^(1/4)]]\)

And by AM-GM again,

\([(a^2 + b^2)^2][sqrt[(c^4 * d^4 * e^4 * f^4)^(1/4)]] >= [(a^2 + b^2)^2][sqrt[(c^2 * d^2 * e^2 * f^2)^(1/2)]]\)

Finally, by AM-GM once more,

\([(a^2 + b^2)^2][sqrt[(c^2 * d^2 * e^2 * f^2)^(1/2)]] >= [(a^2 + b^2)^(1/2)][(c * d * e * f)^(1/2)]^2\)

Applying AM-GM to the expression inside the square root on the right-hand side, we have

\([(a^2 + b^2)^(1/2)][(c * d * e * f)^(1/2)]^2 >= [(a^2 + b^2)^(1/2)][2(abcdef)^(1/6)]^2\)

Finally, squaring both sides, we get

\([(a^2 + b^2)^2][(c * d * e * f)^(1/2)]^2 >= 32abcdef\), as desired.

For part (b), we can use the AM-GM inequality directly:

\((a^2 + b^2)(c^2 + d^2)(e^2 + f^2) = (a^2 + b^2) * [(c^2 + d^2) * (e^2 + f^2)]\)

By AM-GM,

\((a^2 + b^2) * [(c^2 + d^2) * (e^2 + f^2)] >= (a^2 + b^2) * 2sqrt[(c^2 * d^2)(e^2 * f^2)]\)

By AM-GM again,

\((a^2 + b^2) * 2sqrt[(c^2 * d^2)(e^2 * f^2)] >= 2(a^2\)

We can use the formula for average speed, which is the total distance traveled divided by the total time taken. Let's call the total distance traveled "d".

Since Carissa took the same route both ways, the distance traveled uphill is the same in both directions, and the same is true for the distance traveled downhill and across flat ground. So, the total distance traveled uphill is 2d/3, the total distance traveled across flat ground is d/3, and the total distance traveled downhill is 2d/3.

The time taken to travel uphill at 100 meters per hour is 2d/3 * (1 hour/100 meters) = 2d/(300 meters), the time taken to travel across flat ground at 120 meters per hour is d/3 * (1 hour/120 meters) = d/(360 meters), and the time taken to travel downhill at 180 meters per hour is 2d/3 * (1 hour/180 meters) = 2d/540 meters.

The total time taken is 2d/(300 meters) + d/(360 meters) + 2d/540 meters = (12d + 12d + 10d)/(180 meters) = 24d/180 meters = 4d/30 meters.

The average speed during the entire round trip is d/(4d/30 meters) = 30/4 = 7.5 meters per hour.

18 is the right answer.  Thank you!

(a) No, 1/7 is not in the set S.

(b) Yes, 1/4 is in the set S. The number 1/4 can be expressed as 1/3^2 + 1/3^4 + 1/3^5 + ..., which is a series that converges to 1/4.

Try to use Burnside's lemma 

Thanks Mathfreaker.    

I will point out however that It doesn't work if a and b are both 0. 

There needs to be a greater than OR equal to sign.

(a) The left-hand side counts the number of ways to choose k elements from a set of n - 1 elements. The first term, C(n - 1, k - 1), counts the number of ways to choose k - 1 elements, and the second term, C(n - 1, k), counts the number of ways to choose k elements. To form the set of k elements in C(n, k), we simply add an element n to each of these two sets. So, C(n - 1, k - 1) + C(n - 1, k) = C(n, k).

(b) The left-hand side counts the number of ways to choose 0 to n elements from a set of n elements. For each positive integer i from 0 to n, C(n, i) counts the number of ways to choose i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n elements, which is 2^n.

(c) The left-hand side counts the number of ways to choose 0 to k elements from a set of n + k elements. Each term C(n + i, i) counts the number of ways to choose i elements from a set of n + i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n + k elements, which is C(n + k + 1, k).

(d) The left-hand side counts the number of ordered pairs (X, Y) where X is a set of r elements selected from a set of m elements and Y is a set of r elements selected from a set of n elements. For each positive integer i from 0 to r, C(m, i) C(n, r - i) counts the number of ordered pairs where X has i elements and Y has r - i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of r elements from the set of m + n elements, which is C(m + n, r).

(e) The left-hand side counts the number of ordered pairs (a, b) of positive integers where 1 <= a, b <= n. The expression (n + 1)^2 - 2n - 1 counts the total number of ordered pairs of positive integers where 1 <= a, b <= n + 1 and then subtracts the ordered pairs (n + 1, n + 1) and (n + 1, i) for 1 <= i <= n, as well as (i, n + 1) for 1 <= i <= n. This gives us the total number of ordered pairs of positive integers where 1 <= a, b <= n.

(f) The left-hand side counts the number of ordered triples (a, b, c) of positive integers where 1 <= a, b, c <= n. The expression (n + 1)^3 - 3n^2 - 3n - 1 counts the total number of ordered triples of positive integers where 1 <= a, b, c <= n + 1 and then subtracts the ordered triples (n + 1, n + 1, n + 1), (n + 1, n + 1, i), (n + 1, i, n + 1), and (i, n + 1, n + 1) for 1 <= i <= n, as well as all ordered triples where at least one element is equal to n + 1. This gives us the total number of ordered triples of positive integers where 1 <= a, b, c <= n.

This inequality can be proven using the Cauchy-Schwarz inequality.

Let \(x = sqrt(a^2 + ab + b^2), y = sqrt(a^2 + ac + c^2)\), and \(z = sqrt(b^2 + bc + c^2)\), then we can rewrite the inequality as:

\(x + y + z >= sqrt(3) (sqrt(ab) + sqrt(ac) + sqrt(bc))\)

Squaring both sides, we get:

\((x + y + z)^2 >= 3 (sqrt(ab) + sqrt(ac) + sqrt(bc))^2\)

Expanding the left-hand side, we get:

\(x^2 + y^2 + z^2 + 2xy + 2xz + 2yz >= 3 (sqrt(ab) + sqrt(ac) + sqrt(bc))^2\)

Expanding the right-hand side, we get:

\(x^2 + y^2 + z^2 + 2xy + 2xz + 2yz >= 3 (ab + ac + bc + 2 sqrt(ab) sqrt(ac) + 2 sqrt(ab) sqrt(bc) + 2 sqrt(ac) sqrt(bc))\)

Using the Cauchy-Schwarz inequality, we have:

\(2xy + 2xz + 2yz >= 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))\)

Substituting this into the previous inequality, we get:

\(x^2 + y^2 + z^2 + 2xy + 2xz + 2yz >= 3 (ab + ac + bc) + 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))\)

Using the definition of x, y, and z, we get:

\((a^2 + ab + b^2) + (a^2 + ac + c^2) + (b^2 + bc + c^2) + 2 sqrt((a^2 + ab + b^2)(a^2 + ac + c^2)) + 2 sqrt((a^2 + ab + b^2)(b^2 + bc + c^2)) + 2 sqrt((a^2 + ac + c^2)(b^2 + bc + c^2)) >= 3 (ab + ac + bc) + 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))\)

Combining like terms and rearranging, we get:

\(2 (sqrt((a^2 + ab + b^2)(a^2 + ac + c^2)) + sqrt((a^2 + ab + b^2)(b^2 + bc + c^2)) + sqrt((a^2 + ac + c^2)(b^2 + bc + c^2))) >= 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))\)

So, the inequality is proven.

Equality occurs when x = y = z, which means\( a^2 + ab + b^2 = a^2 + ac + c^2 =\)

Yes, there is a theorem that states that\( a^2 + b^2 > ab\) for any real numbers a and b. This theorem is known as the Cauchy-Schwarz inequality.

The Cauchy-Schwarz inequality states that for any two sequences of real numbers \((a_1, a_2, ..., a_n)\) and \((b_1, b_2, ..., b_n)\), we have:

\((a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2) >= (a_1 b_1 + a_2 b_2 + ... + a_n b_n)^2\)

Taking \(n = 2\), we have:

\((a^2 + b^2)(b^2 + b^2) >= (ab + ab)^2\)

Expanding the right-hand side and cancelling the \(b^2\) terms, we get:

\(a^2 + b^2 > ab\)

So, for any real numbers a and b,\( a^2 + b^2 > ab\).

It's a very useful and important theorem in mathematics and has many applications in various fields such as linear algebra, functional analysis, and optimization.

We can use the binomial theorem to find the coefficient of the term \(x^3\) in the expansion of \( (1 + ax)^n\).

The binomial theorem states that, for any real number a and any positive integer n, the expansion of (1 + ax)^n is given by:

\((1 + ax)^n = C(n, 0) * 1^(n-0) * (ax)^0 + C(n, 1) * 1^(n-1) * (ax)^1 + C(n, 2) * 1^(n-2) * (ax)^2 + ... + C(n, n) * 1^(n-n) * (ax)^n\)

where C(n, k) is the binomial coefficient given by:

\(C(n, k) = n! / (k! (n-k)!)\)

where n! means n factorial, \(i.e., n! = n * (n-1) * (n-2) * ... * 1\).

So, to find the coefficient of x^3 in the expansion of \((1 + ax)^n\), we need to find \(C(n, 3) * 1^(n-3) * (ax)^3\).

Since \( 1^(n-3) = 1\), the coefficient of \(x^3\) is simply \(C(n, 3) * ax^3 = n! / (3! (n-3)!) * a^3\).

Therefore, if c is the coefficient of \(x^3\) in the expansion of \((1 + ax)^n\), we have:

\(c = n! / (3! (n-3)!) * a^3\)

So, to find c, we need the values of n and a. Unfortunately, the problem doesn't give us these values, so we cannot find c without more information.

The formula for calculating the balance after t years for an account with an interest rate of r compounded annually is given by:

\(A = P * (1 + r)^t\)

where A is the balance after t years, P is the initial deposit (or principal), and r is the interest rate expressed as a decimal.

For semi-annual compounding, the interest rate is divided by 2, and the formula becomes:

\(A = P * (1 + (r/2))^(2t)\)

For quarterly compounding, the interest rate is divided by 4, and the formula becomes:

\(A = P * (1 + (r/4))^(4t)\)

For monthly compounding, the interest rate is divided by 12, and the formula becomes:

\(A = P * (1 + (r/12))^(12t)\)

Note that in each case, t represents the number of compounding periods (i.e., years for annual compounding, semi-annual periods for semi-annual compounding, quarterly periods for quarterly compounding, and monthly periods for monthly compounding).

So, in your problem, if Chuck deposits $2000 into a bank account that compounds annually at an interest rate of r, the balance after t years will be:

\(A = 2000 * (1 + r)^t\)

If the interest is compounded semi-annually, the balance after t years will be:

\(A = 2000 * (1 + (r/2))^(2t)\)

If the interest is compounded quarterly, the balance after t years will be:

\(A = 2000 * (1 + (r/4))^(4t)\)

If the interest is compounded monthly, the balance after t years will be:

\(A = 2000 * (1 + (r/12))^(12t)\)

Let's call the number of students taking Spanish "x". Then, the number of students taking Chinese is 97 - x.

Since there are more students taking Chinese than Spanish, we know that x < 97 - x, so 2x < 97, which means x < 49.

If x students take only Spanish, then the number of students taking both languages is 97 - x - x = 97 - 2x.

Since x < 49, we know that 97 - 2x > 0, so there are more than 0 students taking both languages.

Therefore, the number of students taking both languages is 97 - 2x, where x is the number of students taking only Spanish.

We can solve this problem using the principle of inclusion-exclusion.

Let's consider the number of ways to color the five squares with no restrictions first. There are 3^5 = 243 ways to color the squares, since each square can be colored in 3 different ways.

Next, let's consider the number of ways to color the squares with the restriction that no two consecutive squares can have the same color. We can use dynamic programming to solve this. Let R[i] be the number of ways to color the first i squares with no two consecutive squares having the same color, such that the i-th square is red. Similarly, let Y[i] and B[i] be the number of ways to color the first i squares with the restriction, such that the i-th square is yellow and blue, respectively. Then, we have:

R[i] = Y[i-1] + B[i-1] Y[i] = R[i-1] + B[i-1] B[i] = R[i-1] + Y[i-1]

Starting with R[1] = 1, Y[1] = 1, B[1] = 1, we can calculate R[2], Y[2], and B[2], and so on, until we reach R[5], Y[5], and B[5]. The final answer is R[5] + Y[5] + B[5].

Applying this dynamic programming approach, we get R[5] = 5, Y[5] = 4, B[5] = 4.

So, there are 5 + 4 + 4 = 13 ways to color the five squares with the restriction that no two consecutive squares can have the same color.

Finally, to find the number of ways to color the squares with the restriction that at least three of the squares are red, we subtract the number of ways to color the squares with fewer than 3 red squares from the number of ways to color the squares with no restrictions.

There are 3 ways to color the squares with 0 red squares, and 3 ways to color the squares with 1 red square, so there are 3 + 3 = 6 ways to color the squares with fewer than 3 red squares.

Therefore, there are 243 - 6 = 237 ways to color the five squares with the restrictions that no two consecutive squares can have the same color and that at least three of the squares are red.

This is a difficult problem, but we can solve it using Burnside's Lemma. Burnside's Lemma is a counting theorem that allows us to count the number of distinct ways of arranging items in a group, while taking into account the symmetry of the group.

First, we need to find the number of fixed points of each possible rotation. A fixed point is a seat that remains unchanged after a rotation.

For a single adult, there are 2 fixed points: the seat directly opposite the adult, and the seat directly to the right of the adult.

For two adults, there are 4 fixed points: the two seats directly opposite the adults, and the two seats directly to the right of the adults.

For three adults, there are 6 fixed points: each of the seats.

So, the number of seatings with no fixed points is 0. The number of seatings with exactly one fixed point is 6 * 2, the number of seatings with exactly two fixed points is 4 * 3, and the number of seatings with exactly three fixed points is 2 * 4.

Finally, we divide the total number of seatings by the number of rotations, which is 6 in this case (since a circular table has 6 rotational symmetries), to get the number of distinct seatings:

(6 * 2 + 4 * 3 + 2 * 4) / 6 = 12 + 12 + 8 / 6 = 32 / 6 = 5.33...

So, there are approximately 5 distinct ways to seat the three adults and three children at the circular table such that each child is next to at least one adult.

a) Each sticker can go to one of five friends, meaning that there are 5^20 ways.