booboo44

At first glance, it might seem like there are 5 options for each side and so a total of 5 * 5 * 5 = 125 ways to color the triangle. However, this overcounts the number of colorings because rotating or reflecting the triangle can create what seems like a different coloring.

We can consider two cases:

Case 1: All three sides are the same color.

There are 5 choices Sam can make for this single color (since all three sides are the same).

Case 2: Two sides are the same color and the third side is a different color.

Here, we need to consider how many ways Sam can choose the two colors and how many ways he can arrange them.

Choosing the colors: There are 5 choices for the color used twice and 4 remaining choices for the different color (since one color is already chosen for the two sides). So, there are 5 * 4 = 20 ways to choose the colors.

Arranging the colors: It might seem like there are 2 ways to arrange the colors (one color for two sides and another for the remaining side). However, rotating the triangle flips the arrangement. So, if we consider both rotations as the same coloring, then there is only 1 way to arrange the colors.

Therefore, for Case 2, the total number of colorings is 20 (choosing colors) * 1 (arranging colors) = 20.

Total Colorings:

Adding the number of colorings for both cases:

Total Colorings = Case 1 + Case 2 = 5 (all same color) + 20 (two same, one different) = 25

So, Sam can color the triangle in 25 different ways.

The sequence 2, 3, 5, 6, 7, 10, 11, ... consists of all positive integers that are neither perfect squares nor perfect cubes.

One way to find the 1000th term is to simply count the number of perfect squares and perfect cubes less than a very large number, then subtract that number from the very large number itself, assuming it is not a perfect square or cube itself.

The 33rd perfect square is 332=1089

The 12th perfect cube is 123=1728

Since 1089<1728, there are more perfect squares than perfect cubes less than 1728.

We know 1 is a perfect square and perfect cube, so we subtract 2 from the count.

Therefore, there are 33−2=31 perfect squares and cubes less than 1728.

Since 1728 is not a perfect square or cube, the 1728th term in Cai's list is 1728.

Following this logic, to find the 1000th term, we can look for the smallest perfect square greater than 1000.

The 32nd perfect square is 322=1024

There are 32−2=30 perfect squares and cubes less than 1024.

Since 1024 is a perfect square, the 1000th term in Cai's list is not 1024. However, we can see that the difference between the 1000th term and the 1024th term is less than or equal to the difference between the 999th term and the 1000th term (which is 1).

Therefore, the 1000th term must be 1 less than the 32nd perfect square, which is 322−1=1023​.

To solve the equation EIGHT / FOUR = TWO, where each letter stands for a different digit, we can start by considering the possible values of the letters.

Since we are dividing a four-digit number by a four-digit number and obtaining a three-digit number, we know that E and F must be 9 and 1, respectively, as they are the leading digits in the division.

With this information, let's construct the division:

```
   9 1
-------
   T W O
```

Now, let's consider the possible values for T and W.

Since 91 divided by 4 is not a whole number, T must be greater than 2. Let's try T = 3:

```
   9 1
-------
  3 W O
```

For the division to yield a three-digit result, the number formed by the last two digits of the dividend (1W) must be divisible by 4. Therefore, W must be an even number. The only even digit left is 2.

```
   9 1
-------
  3 2 O
```

Now, let's find the value of O. Since 91 divided by 4 is 22 with a remainder of 3, O must be 3.

```
   9 1
-------
  3 2 3
```

Therefore, the solution to the equation EIGHT / FOUR = TWO, where each letter stands for a different digit, is:

```
   9 1
-------
  3 2 3
```

So, E = 9, I = 1, G = 3, H = 2, T = 9, W = 2, and O = 3.

Here is the answer to the second problem.

To find the area of the square inscribed in the right triangle, we can use the fact that the square is inscribed, meaning that its diagonal is also the hypotenuse of the right triangle.

Let the side length of the square be \( s \). Since the square is inscribed in the right triangle, its diagonal is equal to the hypotenuse of the right triangle.

By the Pythagorean theorem, the length of the hypotenuse of the right triangle is given by:

\[ \sqrt{1^2 + 3^2} = \sqrt{10} \]

Since the diagonal of the square is also the hypotenuse of the right triangle, its length is \( \sqrt{10} \).

Now, the diagonal of the square divides it into two congruent right triangles. Each right triangle has legs of length \( s \) and \( s \), and a hypotenuse of length \( \sqrt{10} \).

Using the Pythagorean theorem again, we have:

\[ s^2 + s^2 = (\sqrt{10})^2 \]

\[ 2s^2 = 10 \]

\[ s^2 = \frac{10}{2} \]

\[ s^2 = 5 \]

\[ s = \sqrt{5} \]

So, the side length of the square is \( \sqrt{5} \).

The area of the square is the square of its side length:

\[ \text{Area} = (\sqrt{5})^2 \]

\[ \text{Area} = 5 \]

Therefore, the area of the square inscribed in the right triangle is \( 5 \) square units.

Here's how to find the number of times the digit 9 appears in the list of all integers from 1 to 750:

Approach 1: Separating by Hundreds Place

We can break down the count into three sections based on the hundreds digit:

Numbers from 1 to 99:

Every unit digit (from 0 to 9) will appear 10 times (once in each number from 10 to 19, 20 to 29, and so on).

Since 9 is a unit digit, it appears 10 times in this range.

Numbers from 100 to 199:

The hundreds digit is always 1, so we only need to consider the tens and units digits.

Following the same logic as before, each digit (including 9) will appear 10 times in the tens place.

Therefore, 9 appears 10 times in this range.

Numbers from 200 to 750:

We again need to consider both the hundreds and tens/units digits.

The hundreds digit can be 2, 3, ..., 7 (6 different values).

For each hundreds digit value, the tens and units digits will each allow the digit 9 to appear 10 times (as explained earlier).

Counting the Occurrences:

Occurrences in 1-99: 10 times

Occurrences in 100-199: 10 times

Occurrences in 200-750: 6 hundreds digits * 10 occurrences/hundreds digit = 60 times

Total Occurrences:

Adding the occurrences from each section:

Total = 10 (from 1-99) + 10 (from 100-199) + 60 (from 200-750) = 80 times

Approach 2: Utilizing Digit Patterns

Another approach is to analyze the digit patterns within the range.

Single-digit numbers (1-9): The digit 9 appears once.

Two-digit numbers ending in 9 (from 19 to 99): Each number contributes the digit 9 once. There are 10 such numbers (9 total for 10s digit and 1 for the unit digit of 99).

Three-digit numbers with 9 in the tens or units place (from 109 to 759, and from 190 to 799): Similar to two-digit numbers, each number contributes the digit 9 once. There are 10 numbers for each placement of 9 (tens or units), resulting in 20 numbers.

Three-digit numbers with 9 in the hundreds place (from 900 to 999): Each number contributes the digit 9 twice. There are 100 such numbers (10 numbers from 900 to 909 and another 90 numbers from 910 to 999).

Counting the Occurrences:

Single digit: 1 time

Two-digit ending in 9: 10 times

Three-digit with 9 in tens/units: 20 times

Three-digit with 9 in hundreds: 100 times (each digit counted twice)

Total Occurrences:

Adding the occurrences from each pattern:

Total = 1 + 10 + 20 + 2 * 100 = 131 times

Addressing the Discrepancy:

The second approach seems to give a higher count (131) compared to the first approach (80). However, there's a double-counting error in the second approach.

Numbers like 199 and 909 are counted twice (once for 9 in tens place and again for 9 in units place).

Correcting the Count:

We need to subtract the number of times a three-digit number is counted twice for both tens and units digits containing 9.

There are 20 such numbers (as counted earlier).

Final Count:

Total occurrences (corrected) = 131 (initial count) - 20 (double-counted) = 111 times

This corrected count (111) is incorrect. The first approach (80) provides the accurate answer.

The mistake in the second approach highlights the importance of careful analysis and avoiding double-counting when dealing with digit patterns.

Here's how to find the number of blue marbles in the bag:

Total Ratio:

We are given that the ratio of yellow, green, blue, and white marbles is 4:1:2:3. This represents the proportion of each color relative to the total number of marbles.

The sum of these ratios represents the total number of parts that make up the whole bag. In this case:

Total Ratio = 4 + 1 + 2 + 3 = 10

Fraction per Marble Color:

To find the fraction representing the proportion of each color, we can divide each value in the ratio by the total ratio:

Fraction of Yellow = 4 / 10 = 2/5

Fraction of Green = 1 / 10

Fraction of Blue = 2 / 10 (what we're solving for)

Fraction of White = 3 / 10

Total Number of Marbles:

The problem states that the bag contains 20 marbles.

Number of Blue Marbles:

Since the fraction of blue marbles (2/10) represents their proportion of the total number of marbles (20), we can set up a proportion to find the actual number of blue marbles (b):

Fraction of Blue : Total Number of Marbles = Number of Blue Marbles : Total Number of Marbles

Substituting the known values:

(2/10) : 20 = b : 20

Solving for b (Number of Blue Marbles):

Cross-multiplying:

2 * 20 = 10 * b

Divide both sides by 10:

b = 4

Therefore, there are 4 blue marbles in the bag.