blackpanther

Let's analyze the given information and set up an equation to find the desired number of palindromes.

Three-digit palindrome: We can represent the number as abc=100a+10b+c, where a is the hundreds digit, b is the tens digit, and c is the units digit (since it's a palindrome, a=c ).

Sum of digits and product: The problem states that the sum of the hundreds digit, units digit, and the product of the units and tens digits is eight more than the tens digit. Translating this into math, we get: a+c+bc=b+8.

Solving the equation: Since a=c, we can substitute to get: $$ a + a + ab = b + 8.$$ Combining like terms: $$ 2a + ab = b + 8.$$ Factoring out b from the left side: b(a+1)=b+8.

Now we can analyze the solutions for b:

If b=0, then the equation is always true regardless of a. However, a three-digit palindrome cannot have a leading zero (hundreds digit cannot be zero). So, b=0.

If b=0, then we can divide both sides by b to get: a+1=1+b8​. Since a is a digit (0-9), the expression on the right side must be an integer. This is only possible when b is a divisor of 8.

Possible values for b are: b=1,2,4,8. Let's check each case to see if it leads to a valid three-digit palindrome:

b=1: In this case, a=0 (which we cannot have).

b=2: Here, a=3, resulting in the palindrome 323​.

b=4: We get a=1, which leads to the palindrome 141​.

b=8: This results in a=−1, which is not a valid digit.

Therefore, there are only two three-digit palindromes that satisfy the given property: 323 and 141.

Consider the path as a series of 6 decisions (over or under) to be made.

There are 2^6=64 total possible choices, and exactly half of them will result in a knot being formed when the ends are pulled taut.

Therefore, the probability is 1/2.

Here's the solution for 1.

We can approach this problem by analyzing the given inequality and maximizing the expression for x^2 + 5x + 6 within the constraints.

Factoring the Inequality: The inequality x^2 + 7x + 12 <= 0 can be factored as (x + 3)(x + 4) <= 0. This means either both factors are less than or equal to zero, or one factor is positive and the other is negative (the product becomes zero).

Identifying the Range of x: From the factored inequality, we know:

x ≤ -3 (when x + 3 is less than or equal to zero)

x ≤ -4 (when x + 4 is less than or equal to zero)

However, the second condition (x ≤ -4) is redundant as -3 is already less than or equal to -4. Therefore, the valid range for x is x ≤ -3.

Maximizing x^2 + 5x + 6: We want to maximize the value of x^2 + 5x + 6 within the constraint x ≤ -3. We can rewrite the expression as a completed square:

x^2 + 5x + 6 = (x^2 + 5x + 25/4) + 11/4 = (x + 5/2)^2 + 11/4

Since the square of any real number (x + 5/2) is non-negative (including zero), adding a constant positive value (11/4) will always result in a positive or zero value.

Largest Possible Value: The largest possible value for x^2 + 5x + 6 occurs when (x + 5/2)^2 is zero. This happens when x = -5/2. However, this value of x (-5/2) violates the constraint x ≤ -3.

Therefore, the largest possible value for x^2 + 5x + 6 within the allowed range of x is achieved when x is as close to -3 as possible (without exceeding it). This occurs when x = -3.

Evaluating the Maximum Value: Substituting x = -3 in the expression:

x^2 + 5x + 6 = (-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0

Answer: The largest possible value of x^2 + 5x + 6 is 0.

Let m=37. We can show that n is divisible by 37 if and only if the result of this process is divisible by 37 as follows:

If n is divisible by 37, then the alternating sum of the two-digit chunks is divisible by 37.

This is because the alternating sum of the two-digit chunks is just the sum of the digits of n, modulo 37. Since n is divisible by 37, the sum of its digits is also divisible by 37, so the alternating sum of the two-digit chunks is also divisible by 37.

If the alternating sum of the two-digit chunks is divisible by 37, then n is divisible by 37.

This is because the sum of the digits of n is congruent to the alternating sum of the two-digit chunks, modulo 37.

Therefore, if the alternating sum of the two-digit chunks is divisible by 37, then the sum of the digits of n is also divisible by 37, so n is divisible by 37.

Therefore, the alternating sum of the two-digit chunks is a divisibility test for 37.

To see this in action, let's consider the number 354764. The alternating sum of the two-digit chunks of 354764 is 64−47+35=52. Since 52 is divisible by 37, we know that 354764 is also divisible by 37.

There are three cases to consider, since all three marbles can be red, all white, or all blue.

Case 1: All Red Marbles

Favorable cases: We need to choose 3 red marbles from 4 available. This can be done in ⁴C₃ ways (combinations, not permutations).

Total cases: We can choose any 3 marbles from a total of 4 red + 5 white + 6 blue = 15 marbles in ¹⁵C₃ ways.

Case 2: All White Marbles

Favorable cases: We need to choose 3 white marbles from 5 available. This can be done in ⁵C₃ ways.

Total cases: Same as Case 1 (¹⁵C₃).

Case 3: All Blue Marbles

Favorable cases: We need to choose 3 blue marbles from 6 available. This can be done in ⁶C₃ ways.

Total cases: Same as Case 1 (¹⁵C₃).

Total Probability

The probability that all three marbles are the same color (considering all three cases) is the sum of the probabilities of each case:

P(all same) = P(all red) + P(all white) + P(all blue)

Calculate Each Probability:

P(all red) = ⁴C₃ / ¹⁵C₃ P(all white) = ⁵C₃ / ¹⁵C₃ P(all blue) = ⁶C₃ / ¹⁵C₃

We can simplify each term using the fact that nCr = n! / (r! * (n-r)!):

P(all red) = (4!) / ((3!) * (1!)) / ((15!) / ((3!) * (12!))) P(all white) = (5!) / ((3!) * (2!)) / ((15!) / ((3!) * (12!))) P(all blue) = (6!) / ((3!) * (3!)) / ((15!) / ((3!) * (12!)))

Notice that (3!) and (12!) terms appear in all the denominators and numerators (except the first factorials in the numerators). These terms will cancel out, leaving:

P(all red) = 4 / (15 * 14 * 13) P(all white) = 10 / (15 * 14 * 13) P(all blue) = 20 / (15 * 14 * 13)

Final Probability:

Add the probabilities of each case:

P(all same) = 4/210 + 10/210 + 20/210 = 34/210

Simplify the fraction:

P(all same) = 17/105

Therefore, the probability that all three marbles drawn are the same color is 17/105.

To find the probability of winning a super prize in the SuperLottery, we need to consider two mutually exclusive events:

1. **Winning by matching at least two of the white balls.**

2. **Winning by matching the red SuperBall.**

We'll calculate the probabilities for each event and then add them together.

1. **Winning by matching at least two of the white balls:**

   To calculate this probability, we can find the probability of not matching any of the white balls and subtract it from 1.

   The probability of not matching any of the white balls on a single draw is:

   \[\frac{{9 \choose 3}}{{12 \choose 3}}\]

   So, the probability of matching at least two of the white balls is:

   \[1 - \frac{{9 \choose 3}}{{12 \choose 3}}\]

2. **Winning by matching the red SuperBall:**

   The probability of matching the red SuperBall is simply \( \frac{1}{8} \) since there's only one SuperBall drawn from 8 possibilities.

Now, let's calculate these probabilities:

1. Probability of winning by matching at least two of the white balls:

   \[1 - \frac{{9 \choose 3}}{{12 \choose 3}} = 1 - \frac{84}{220} = 1 - \frac{21}{55} = \frac{34}{55}\]

2. Probability of winning by matching the red SuperBall:

   \[P(\text{Red SuperBall}) = \frac{1}{8}\]

Finally, to find the total probability of winning a super prize, we add the probabilities of the two mutually exclusive events:

\[P(\text{Winning super prize}) = P(\text{White balls}) + P(\text{Red SuperBall}) = \frac{34}{55} + \frac{1}{8}\]

\[= \frac{272}{440} + \frac{55}{440}\]

\[= \frac{272 + 55}{440}\]

\[= \frac{327}{440}\]

So, the probability of winning a super prize in the SuperLottery is \( \frac{327}{440} \).

Since PT and PU are tangent to the circle centered at O, then ∠PTO and ∠PUO are right angles. Additionally, since ∠TOP=45∘, then △PTO is a 45-45-90 triangle.

In a 45-45-90 triangle, each of the acute angles measures 45∘, and the hypotenuse (opposite the right angle) has a length 2​ times the length of either leg. Therefore, TO=OP⋅2​.

Because TU is a minor arc and ∠TOP=45∘, then arc TU intercepts a central angle ∠TOU that also measures 45∘.

Since central angles and their corresponding minor arcs have the same measure in degrees, then minor arc TU has a measure of 45∘​.

Since triangle ABC is isosceles with AB=AC=20, segment AM is a median and an altitude (since the base angles are congruent). Therefore, △ABM is a right triangle with hypotenuse AB=20 and leg BM=21​⋅AB=10

Because D is the midpoint of AM, MD=21​⋅AM=21​⋅AB=10. Then △BDM is a 3-4-5 right triangle, so BD=4⋅10=40.

Similarly, since E is the midpoint of AB, AE=21​⋅AB=10. Then △AEM is also a 3-4-5 right triangle, so ME=4⋅10=40.

Triangle BET shares altitude EM with △AEM, and the ratio of their bases is AEBE​=21​. Therefore, the area of △BET is half the area of △AEM, which is 21​⋅21​⋅AE⋅EM=21​⋅21​⋅10⋅40=100​.