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The answer is \(2\)

We must take the highest powers and add them together. 

\(5+6=11\)

So the answer is 11. 

Thanks! :)

Ok, we can go one by one to solve these problems. 

Hoever, let's note that if we have f(x) = x, then we set the right side of the function to equal x.

Then, we see if the x value satisfies the conditions given. 

First, we have \(f(x) = 2x + 8\). Setting this to equal x, we find that \(x = 2x + 8\)

Solving for x, we find that \(x=-8\). This doesn't satisfy the condition \(1 \le x \le 2\), so it is not a valid solution. 

Next, we have \(f(x) = 13 - 5x\). Setting this to equal x, we have \(x = 13 - 5x\)

Finding the value of x, we get \(x=13/6\). This satisfies the inerval, so it is a solution. 

Third, we have \(f(x) = 20 - 14x\). Since this equals x, we can write \(x= 20 - 14x\)

Trying to find x, we have that \(x=20/15=4/3\). This does NOT satisfy the interval, so it is not a solution. 

Lastly, we have \(f(x) = 40 + 5x\). setting this function to equal x, we have \(x = 40 + 5x\)

Solving for x, we get \(x=-10\), which is invalid. 

Thus, the only value that works is 13/6. 

So our answer is 13/6. 

Thanks! :)

Ok, so we start off with the equation

\(ab -3a + 4b = 131\)

I know this is random, but let'st ake the product of the coefficients on a,b.....add this product to  both sides

So this gets us

\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \\ ab - 3a + 4b - 12 = 119\)

This eventually achieves the equation

\((a + 4) ( b - 3) = 119\)

Now, the factors of 119 are \(1,17,119\)

Since we want the minimum, let's use 7 and 17. We have

\(( 13 + 4) ( 10 - 3) → | a - b | = |13 - 10 | = 3\)

Thus, our final answer is 3. 

Thanks! :)

Let's note that If AD  = AB  then angle  ABD  =  angle ADB

This means that \(\angle ABD  = (180 - 17)  / 2   =  81.5°  \)

So 81.5 is the answer. 

Thanks! :)

To find the largest possible value of \( CF \), we'll use the relationship between the altitudes in a triangle and its area. The area \( \Delta \) of a triangle can be expressed using any of its altitudes:

\[
\Delta = \frac{1}{2} \times \text{base} \times \text{height}
\]

For triangle \( ABC \), we have the following relationships:

- \( \Delta = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times CA \times BE = \frac{1}{2} \times AB \times CF \)

Let the side lengths be \( a = BC \), \( b = CA \), and \( c = AB \). Then:

\[
\Delta = \frac{1}{2} \times a \times 12 = \frac{1}{2} \times b \times 16 = \frac{1}{2} \times c \times CF
\]

This simplifies to:

\[
\Delta = 6a = 8b = \frac{1}{2} c \times CF
\]

Equating these expressions:

\[
6a = 8b \quad \text{and} \quad 6a = \frac{1}{2} c \times CF
\]

### Step 1: Solve for \( a \) and \( b \)

From \( 6a = 8b \):

\[
\frac{a}{b} = \frac{8}{6} = \frac{4}{3}
\]

Thus, \( a = \frac{4}{3}b \).

### Step 2: Substitute into \( 6a = \frac{1}{2} c \times CF \)

Substitute \( a = \frac{4}{3}b \) into \( 6a = \frac{1}{2} c \times CF \):

\[
6 \times \frac{4}{3}b = \frac{1}{2} c \times CF
\]

Simplifying:

\[
8b = \frac{1}{2} c \times CF \quad \text{so} \quad 16b = c \times CF
\]

### Step 3: Find the Maximum Value of \( CF \)

We need to maximize \( CF \), which is a positive integer. Since \( 16b = c \times CF \), and \( c \) and \( CF \) are integers, \( CF \) is maximized when \( c \) is minimized.

Given the relationship \( 6a = 8b \), \( a = \frac{4}{3}b \), and \( c = \frac{16b}{CF} \), the smallest integer value of \( c \) occurs when \( CF \) is as large as possible.

If \( CF = 16 \), then:

\[
16b = c \times 16 \quad \text{so} \quad c = b
\]

Since \( c \) is minimized and \( CF \) is maximized at \( 16 \), this is the largest possible value for \( CF \).

Thus, the largest possible value of \( CF \) is \( \boxed{24} \).

To solve this problem, we need to understand the relationships within the triangle STU. Here are the steps to find the length of \( SX \):

### Step 1: Analyzing the Triangle

Given:

- \( S \), \( T \), and \( U \) are the vertices of the triangle.

- \( M \) is the midpoint of \( ST \).

- \( N \) is a point on \( TU \) such that \( SN \) is the altitude of the triangle.

- \( ST = SU = 13 \), \( TU = 8 \).

- \( UM \) and \( SN \) intersect at \( X \).

### Step 2: Applying the Median and Altitude Properties

Since \( M \) is the midpoint of \( ST \), \( SM = MT = \frac{13}{2} = 6.5 \).

Also, \( SN \) is an altitude, so it is perpendicular to \( TU \).

### Step 3: Use the Property of the Centroid

In any triangle, the centroid (intersection of the medians) divides each median in a 2:1 ratio. Since \( X \) is the intersection of the medians \( SN \) and \( UM \), it is the centroid of triangle \( STU \).

This implies:

\[
SX = \frac{2}{3} \times SN
\]

where \( SN \) is the altitude from \( S \) to \( TU \).

### Step 4: Calculate SN Using the Area of the Triangle

We use the fact that the area of the triangle can be calculated in two ways:

1. Using base \( TU \) and height \( SN \).

2. Using Heron's formula.

#### Heron's Formula:

First, calculate the semi-perimeter \( s \):

\[
s = \frac{ST + SU + TU}{2} = \frac{13 + 13 + 8}{2} = 17
\]

Then, calculate the area \( \Delta \):

\[
\Delta = \sqrt{s(s - ST)(s - SU)(s - TU)} = \sqrt{17(17 - 13)(17 - 13)(17 - 8)} = \sqrt{17 \times 4 \times 4 \times 9} = \sqrt{2448} = 24
\]

#### Area Using Altitude \( SN \):

The area can also be written as:

\[
\Delta = \frac{1}{2} \times TU \times SN = \frac{1}{2} \times 8 \times SN = 4 \times SN
\]

Equating the two expressions for the area:

\[
24 = 4 \times SN \implies SN = 6
\]

### Step 5: Calculate SX

Now that we know \( SN = 6 \), the length of \( SX \) is:

\[
SX = \frac{2}{3} \times 6 = 4
\]

Thus, the length of \( SX \) is \( \boxed{4} \).

N mod 10 = a

N mod 2 = b

N mod 20 =?

2 and 10 are not coprime. 10 is multiple of 2.

If a is odd then b=1.

If a is even then b=0.

We don’t need b.

N mod 20 ={a, a+10}

Using modular arithmetic, we find

2^2014 - 1 == 1 (mod 2)

2^2014 - 1 == 1 (mod 3)

2^2014 - 1 == 0 (mod 5)

2^2014 - 1 == 0 (mod 7)

So the product of the two smallest prime factors is 5*7 = 35.

Analyzing the Problem

To return home after exactly 8 blocks, Gracie must walk 4 blocks east and 4 blocks west. The order of these blocks doesn't matter as long as she ends up at home.

Calculating the Probability

Total possible outcomes: There are 2 possible outcomes (heads or tails) for each of the 8 blocks, so there are 2^8 = 256 total possible outcomes.

Favorable outcomes: We need to choose 4 out of the 8 blocks to be eastbound (the remaining 4 will be westbound). This is a combination problem. The number of ways to choose 4 blocks out of 8 is 8C4 = 70.

Calculating the Probability

Probability = Favorable outcomes / Total possible outcomes

Probability = 70 / 256 = 35/128

Therefore, the probability that Gracie stops her walk after walking exactly eight blocks is 35/128.

Thank you all for the help! Sadly, both 5 and 14 were wrong, and the right answer was 15 lol. Here's the problem explanation that I got after getting the problem wrong: 

We use Simon's Favorite Factoring Trick: observe that \(ab+2a+3b=a(b+2)+3b\), so to obtain another factor of \(b+2\), we need to add \(6\). So \(ab+2a+3b+6=a(b+2)+3(b+2)=(a+3)(b+2)\), and \(ab+2a+3b=(a+3)(b+2)-6\).

Thus we need to find how many integers \(n\) with \(76\le n\le96\) can be written as \((a+3)(b+2)\) for at least one ordered pair of positive integers \((a,b)\).    Any \(n\) that works must have the property that there exists positive integers \(n_1\) and \(n_2\) with \(n_1n_2=n\) and \(n_1>n_2 \ge 3\). Otherwise, we wouldn't be able to satisfy the condition that \(a\) and \(b\) are positive integers. That is, \(n=78\) works because \(78 = 26 \cdot 3 = (23 + 3)(1 + 2)\), but \(n = 82\) doesn't because the only possible ways to decompose \(82\) as a product of two positive integers are \(41 \cdot 2\) and \(82 \cdot 1\), and in neither of those cases can both \(a\) and \(b\) be positive integers.

The rest is a brute force check: we need to eliminate prime numbers, as they definitely don't allow us to have \(a\) and \(b\) be positive integers, and we need to eliminate numbers of the form \(2p\), where \(p\) is prime. Every other number has a pair of factors \(n_1\) and \(n_2\) that satisfy \(n_1n_2=n\) and \(n_1>n_2 \ge 3\).

We complementary count: \(79, 83, 89\) are primes, and \(82, 86, 94\) are twice a prime. Our answer is thus the total number of integers \(n\) between \(76\) and \(96\) minus \(6\), or simply \(\boxed{15}\).

To solve for the number of integers \( n \) within the interval \( 70 \leq n \leq 90 \) that can be expressed in the form

\[
n = ab + 2a + 3b,
\]

we can rearrange the equation for \( n \):

\[
n = ab + 2a + 3b = a(b + 2) + 3b.
\]

Now, we can let \( m = b + 2 \). Then, we can rewrite \( b \) in terms of \( m \) as \( b = m - 2 \). Substituting this back into our formula gives:

\[
n = a(m) + 3(m - 2) = am + 3m - 6 = (a + 3)m - 6.
\]

Now, we can rearrange this to isolate \( m \):

\[
n + 6 = (a + 3)m \implies m = \frac{n + 6}{a + 3}.
\]

Since \( m \) is a positive integer, \( n + 6 \) must be divisible by \( a + 3 \). We can explore which integers give permissible \( m \) values by determining values of \( n + 6 \):

- The smallest \( n \) is \( 70 \), so \( n + 6 = 76 \).

- The largest \( n \) is \( 90 \), so \( n + 6 = 96 \).

Consequently, we need to examine the integers from \( 76 \) to \( 96 \):

\[
76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96.
\]

Next, we can determine the possible values of \( a + 3 \). Since \( a \) is a positive integer, \( a \geq 1 \) implies \( a + 3 \geq 4 \).

Therefore, \( m \) can take any integer value for \( a + 3 \in \{4, 5, 6, \ldots\}\).

To find integers \( n \), we need \( n + 6 \) to be divisible by each \( a + 3 \):

- For \( k = 4 \): \( n + 6 = 76 \), \( n \equiv 0 \mod 4 \) gives \( n = 70, 74, 78, 82, 86, 90 \).

- For \( k = 5 \): \( n + 6 = 76 \equiv 1 \mod 5 \) means \( n \equiv -1 \mod 5 \), giving \( n = 74, 79, 84, 89 \).

- For \( k = 6 \): \( n + 6 = 76 \equiv 4 \mod 6 \) means \( n \equiv 2 \mod 6 \), yielding \( n = 72, 78, 84, 90 \).

- For \( k = 7 \): \( n + 6 = 76 \equiv 6 \mod 7 \) which gives \( n \equiv 1 \mod 7 \) as \( n = 70, 77, 84, 91 \).

- For \( k = 8 \): \( n + 6 = 76 \equiv 4 \mod 8 \) gives \( n \equiv 2 \mod 8 \), yielding \( n = 70, 78, 86 \).

- For \( k = 9 \): \( n + 6 = 76 \equiv 4 \mod 9 \) gives \( n \equiv 5 \mod 9 \), yielding \( n = 74, 83, 92 \).

Now, we can collect all the possible \( n \):

From \( k = 4 \): \( 70, 74, 78, 82, 86, 90 \)

From \( k = 5 \): \( 74, 79, 84, 89 \)

From \( k = 6 \): \( 72, 78, 84, 90 \)

From \( k = 7 \): \( 70, 77, 84, 91 \)

From \( k = 8 \): \( 70, 78, 86 \)

From \( k = 9 \): \( 74, 83, 92 \)

Next, let's find unique \( n \) from all these lists:

- Compiling unique values from the sets we found: \( 70, 72, 74, 77, 78, 79, 82, 83, 84, 86, 89, 90, 91, 92 \).

Thus, the unique values of \( n \) between \( 70 \leq n \leq 90 \) are:

\[
70, 72, 74, 77, 78, 79, 82, 83, 84, 86, 89, 90.
\]

Counting these gives us:

\[
70, 72, 74, 77, 78, 79, 82, 83, 84, 86, 89, 90 \Rightarrow \text{ 12 integers. }
\]

Therefore, the number of integers \( n \) that can be expressed in the desired form is \( \boxed{12} \).

Here, let me give this problem a shot. 

First, let's try to simplify this problem a bit. We can use Simon's Favorite Factoring Trick. 

Let's factor the right side a bit for n. We get

\(n = ab + 2a + 3b \\ n = a(b + 2) + 3b \)

Now, since we want b to be factored as well, let's add 6 to both sides. This will come in handy later. 

We get

\( n + 6 = a(b + 2) + 3(b + 2) \\ n +6 = (a + 3)(b + 2)\)

Now, this problem has transformed for numbers that match the conditions

\(76\leq (a+3)(b+2) \leq 96\)

The only numbers that don;t satsify these conditions are numbers with only two factors and one is less than 3. Thus, we can calculate each number. Here, I did it for you! :)

Thus, since there are a total of 21 numbers and 7 are not valid, we have 21 - 7 = 14 as our final answer. 

So 14 should be the answer. I'm not confident on the answer, I might have errored a bit. However, I think it's CLOSE to the right answer. Maybe 12 or 15, but 14 should be right. 

Thanks! :)

Thank you so much for your help, but that didn't work... I'm wondering where you got the \(-6\) from in "n = a(b + 2) + 3(b + 2) - 6", and if that might be the problem? Could it be that you added it to one side, but not the other?