bingboy

The key to this problem lies in the fact that Cindy can have 22 different pile sizes (Y).

Here's why this information helps us find the smallest number of coins:

Divisibility: Since Cindy can have 22 different pile sizes, the total number of coins (let's call it T) must be divisible by 22 different numbers.

Smallest Divisors: The smaller the pile sizes (Y), the more divisors T will have. In other words, for T to have the smallest number of coins, we want the first 22 positive integers to divide T.

Smallest Multiple: The smallest number of coins (T) that satisfies all these conditions will be the least common multiple (LCM) of the first 22 positive integers.

Finding the LCM of all 22 numbers can be computationally expensive. However, we can make some observations:

All the first 11 positive integers divide any larger positive integer.

We only need to focus on the remaining prime numbers less than or equal to 22 (which are 2, 3, 5, 7, 11, 13, 17, and 19).

Checking Divisibility: We can check if T is divisible by each of these prime numbers. If it's not, we need to increase T until it becomes divisible by all of them.

Finding the Smallest T:

Start with a small number like T = 22 (which is divisible by 2 and 11). Check if it's divisible by 3, 5, 7, 13, 17, and 19. If not, increase T by 1 and repeat the checks.

The smallest T that satisfies all the divisibility conditions is T = 88.

Therefore, the smallest number of coins Cindy could have is 88​.

Let's analyze each condition to see what it tells us about the minimum number of marbles in the box.

Red Marbles: Lamant needs at least 20 marbles to guarantee at least one red marble. This means there could be a box with only 19 non-red marbles (blue and green combined).

Green Marbles: Similarly, at least 24 marbles are needed to guarantee at least one green marble. This implies there could be a box with only 23 non-green marbles (red and blue combined).

All Three Colors: Finally, 27 marbles are needed to ensure all three colors are present. This means there could be a box with only 26 marbles that don't have all three colors (a combination of red and blue only, or blue and green only, or red and green only).

Combining the Information:

Looking at these conditions, we see that the number of non-red marbles (blue and green combined) must be greater than 19 (from condition 1) but less than or equal to 23 (from condition 2).

Similarly, the number of non-green marbles (red and blue combined) must be greater than 23 (from condition 2) but less than or equal to 26 (from condition 3).

The only way to satisfy both constraints is if there are exactly 23 non-red marbles and exactly 23 non-green marbles. This means the box must have a total of:

Red marbles = Total - Non-red marbles = Total - 23

Green marbles = Total - Non-green marbles = Total - 23

Finding the Total:

Since the number of red and green marbles must be the same (both equal to Total - 23), the total number of marbles in the box (including red, blue, and green) must be a multiple of 2. The smallest multiple of 2 that satisfies both conditions is:

Total = Red marbles + Green marbles + Blue marbles

Total = (Total - 23) + (Total - 23) + Blue marbles

2 * Total = 46 + Blue marbles

Since the total needs to be a whole number, the number of blue marbles must be even (because multiplying by 2 gives an even number). The smallest even number that satisfies all the conditions is:

Blue marbles = 2

Therefore, the total number of marbles in the box is:

Total = (46 + Blue marbles) / 2

Total = (46 + 2) / 2

Total = 48 / 2

Lamant's box contains a total of 24​ marbles.

Here's how to find [BP C] [ABC] · [BMC] [ABC]:

Relate Areas using Tangent Lengths:

Since the incircle is tangent to the sides of the triangle at points M, N, and P, we know that AM, BN, and CP are radii of the incircle. Let the radius of the incircle be r.

The area of a triangle can be expressed using its base and height:

Area of ∆ABM = (AB)(AM)/2

Area of ∆BCN = (BC)(BN)/2

Area of ∆ACP = (AC)(CP)/2

Relate Base and Height using Angle Properties:

Since angle BAC = 30°, triangle ABC is a 30-60-90 triangle. This means:

AB = 2 * AC (where AC is the base of ∆ABM)

BC = AC√3 (where BC is the base of ∆BCN)

Substitute and Simplify:

Now, substitute the expressions for the bases from step 2 into the area equations from step 1:

Area of ∆ABM = (2 * AC)(r)/2 = AC * r

Area of ∆BCN = (AC√3)(r)/2 = (AC√3 * r)/2

Area Ratio:

We are interested in the ratio of the areas of triangles BPC and BMC:

[BPC] / [BMC] = (Area of ∆BCN) / (Area of ∆ABM) = [(AC√3 * r)/2] / (AC * r) = √3 / 2

Double Counting and Final Answer:

Notice that the area of triangle ABC appears twice in the product we need to find: once in the numerator ([BP C] [ABC]) and once in the denominator ([BMC] [ABC]).

This is because both triangles BPC and BMC share a base (BC) and height (altitude from A to BC) with triangle ABC.

Since the area of triangle ABC cancels out, we are left with:

[BP C] [ABC] · [BMC] [ABC] = √3 / 2 * √3 / 2 = (3/4)

Therefore, the product [BP C] [ABC] · [BMC] [ABC] is equal to 3/4.

Let's track the amount of water in the beaker throughout the process.  Initially, there are 200 mL of water.

First drainage and addition:

Ruth drains 200 mL of solution, which includes some water. Since the solution is well mixed, the proportion of water drained will be the same as the proportion of water in the whole solution.

Proportion of water = 200 mL water / (800 mL acid + 200 mL water) = 1/5

Therefore, the amount of water drained is 1/5 * 200 mL = 40 mL.

After draining, there's 200 mL water - 40 mL drained water = 160 mL water remaining.

Then, Ruth adds 200 mL of pure water, bringing the total water content to 160 mL + 200 mL added water = 360 mL.

Second drainage and addition:

Following the same logic as the first drainage, 40 mL of water gets drained out of the 200 mL solution removed (1/5 proportion).

There's 360 mL water - 40 mL drained water = 320 mL water remaining.

Finally, adding 200 mL more water brings the final amount of water to 320 mL + 200 mL added water = 520 mL.

Therefore, there are now 520 mL of water in the beaker.

Since the arrangement is circular, any specific order will only be unique up to a rotation. So, choosing any person as the "starting point" and fixing them in a seat won't actually affect the number of possible arrangements.

Let's consider one adult, A. For A to be next to two students, there must be two students flanking A. There are 4 choices we can make for the first student to sit next to A.

Once we've chosen that student, there are 3 students left, and only one of them can sit on the other side of A. So, there are 4⋅1=4 ways to arrange the students such that A is flanked by two students.

Now, we have 3 adults left to seat. Since they each must be flanked by two students as well, we can repeat the same process for each adult, giving us 4⋅4⋅4=64​ total arrangements.

We can solve the problem as follows.

Similar Triangles:

Since AB = AC and altitude AH intersects BC at D, we can split triangle ABC into two congruent right triangles (ABH and ACH).

Pythagorean Theorem in ABH:

We are given AH = 11 and DH = 5. Since these sides form a right angle (angle AHB = 90°), we can use the Pythagorean Theorem:

AB^2 = AH^2 + HD^2

AB^2 = 11^2 + 5^2

AB^2 = 121 + 25

AB^2 = 146

Taking the square root of both sides (remembering AB could be positive or negative, but the area is based on the absolute value):

AB = √146 (positive square root for finding side length)

Area of Triangle ABC:

Since triangle ABC is formed by two congruent right triangles (ABH and ACH), their areas will be equal.

Therefore, the area of triangle ABC is simply twice the area of triangle ABH.

Area of triangle ABH = 1/2 * base * height = 1/2 * AB * AH

Substitute Known Values and Simplify:

We know AH = 11 and from step 2, AB = √146. Substitute these values:

Area of triangle ABC = 2 * (1/2 * √146 * 11)

Therefore, the area of triangle ABC is 11*sqrt(146).

Here is a full solution.

Analyze the given information:

Point P is on side AC of triangle ABC.

Angles ACB and ABC are congruent (angle ACB = angle ABC).

The difference between angle ABC and angle ACB is 42 degrees (angle ABC - angle ACB = 42°).

Identify key properties of triangles:

The sum of angles in a triangle is always 180 degrees.

Solve for missing angle in triangle ABC:

Let x represent the unknown measure of angle ABC (and angle ACB by congruence).

From the triangle property: x + x + (x - 42) = 180°

Combine like terms: 2x - 42 = 180

Add 42 to both sides: 2x = 222

Divide both sides by 2 to find x: x = 111°

Find angle PBC:

Since angles ABC and ACB are congruent, each measures 111 degrees (as found previously).

Angle PBC is supplementary to angle ABC (they share a straight line).

Therefore, angle PBC = 180° - angle ABC = 180° - 111° = 69°.

Answer: Angle PBC = 69 degrees.

Properties of Equilateral Triangle:

Since triangle DEF is equilateral, all three sides (DE, DF, and EF) are equal in length. Additionally, all three angles are also equal to 60 degrees.

Right Triangle ACB:

Given that angle ACB = 90 degrees and CD = 5 and CE = 4, we recognize this as a Pythagorean triple (3, 4, 5). Therefore, AC = 3.

Isosceles Triangle DEC:

Since DE = DC + CE and all sides of triangle DEF are equal, triangle DEC is isosceles with DE = 9.

Dropping an Altitude from A:

Draw an altitude from point A to side DE, intersecting DE at point H. Since triangle ACB is a right triangle with a 90-degree angle at C,

line segment AH is also an altitude of triangle DEC, splitting it into two congruent right triangles (ADH and CHE).

Finding AH:

In right triangle ADH, we know the hypotenuse (AD = AC = 3) and want to find the altitude (AH).

Since angle A is a vertex of an equilateral triangle, angle HAD = 60 degrees (half of an equilateral triangle's angle).

We can use the trigonometric ratio tangent (tan) for this situation: tan(HAD) = AH/HD.

Knowing tan(60) = √3 and that HD = DE/2 (since AH bisects DE), we can set up the equation: tan(60) = √3 = AH / (DE/2)

Solving for AH: AH = √3 * (DE/2) = √3 * (9/2) = 3√3 / 2

Finding AE:

In right triangle AEH, we know the altitude (AH = 3√3 / 2) and the base (HE = CE - CH = 4 - AC = 4 - 3 = 1).

We can use the Pythagorean theorem: AE^2 = AH^2 + HE^2

Substitute the known values: AE^2 = (3√3 / 2)^2 + 1^2 = 27/4 + 1 = 31/4

Take the square root of both sides to find AE: AE = sqrt(31)/2

Therefore, the length of AE is equal to sqrt(31)/2.

The system

x + y + 3z = 10,
-4x + 2y + 5z = 7,
kx + z = 3

will have no solution if the corresponding matrix

[1 1 3]
[-4 2 5]
[k 0 1]

has a determinant of 0.  Taking determinant, we find that it is 2k - 14.

Therefore, the value of k such that the system has no solution is k = 7.

Absolutely, I've been improving my problem-solving abilities in counting problems. Let's tackle this step-by-step.

(a)  (Observe the digits of the numbers 1 to 250).

Numbers from 1 to 9 have 1 digit each (9 numbers).

Numbers from 10 to 99 have 2 digits each (90 numbers).

Numbers from 100 to 250 have 3 digits each (151 numbers).

Therefore, the total number of digits is 9×1+90×2+151×3=9+180+453=642.

(b) (Calculate the number of times the digit 2 appears)

Each number from 10 to 19 contains the digit 2 once. (10 numbers)

Each number from 20 to 29 contains the digit 2 once. (10 numbers)

Similarly, each hundred from 100 to 240 contains the digit 2 ten times (15 hundreds * 10 = 150 times)

Additionally, numbers 200, 210, 220, 230, and 240 each contain the digit 2 twice. (5 numbers * 2 = 10 times)

So, the total number of times the digit 2 appears is 10+10+150+10=180.

(c)  The probability of Sam choosing a 2 is the number of times 2 appears divided by the total number of digits:

Probability = (Number of 2s) / (Total number of digits) = 180 / 642 = 20/71

Therefore, the probability of Sam choosing a 2 is 20/71.