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Think!

This is not a hard one.

The answer is 3.

This is because if one term is in the form of an x^2, the lowest power needs to be 2. However, since we also have a y, we know the equation can't be a quadratic (think it through!), and therefore the only power sufficing this form would be 3.

Does that make sense?

Assumig peope are the same an rotations are the same

then

tie each person in a chair  to a chair  on the right,  

so now there are 5 double people and 8 other seats that is 13 altogether

Pur one double in place, down't matter where, that takes care of rotations

now there are    12C4 = 495 ways

are the people classed as identical?

are rotations the same?

This is your aops teacher and I want to tell you that we will have a meeting seperately the next class. I want to let you know that this is unacceptable as I will be emailing your parents about your actions. 

There are four ways in which a mailbox could have exactly 3 letters: the first mailbox could have 3 letters, the second mailbox could have 3 letters, the third mailbox could have 3 letters, or the fourth mailbox could have 3 letters.

Let's consider each case separately:

Case 1: The first mailbox has 3 letters.

In this case, there are 3 letters left to be placed in the remaining 3 mailboxes. Each of the remaining 3 letters can be placed in any of the 3 remaining mailboxes, so there are 3 choices for each of the 3 remaining letters. Therefore, there are 3^3 = 27 ways in which the remaining letters could be placed in the remaining mailboxes.

Case 2: The second mailbox has 3 letters.

This case is similar to Case 1, but we need to account for the fact that the 3 letters that go in the second mailbox could be any 3 of the 4 letters. There are 4 ways to choose which 3 letters go in the second mailbox, and then there are 3 choices for the remaining letter, and 2 choices for where to put it. Therefore, there are 4 x 3 x 2 = 24 ways in which the letters could be placed in this case.

Case 3: The third mailbox has 3 letters.

This case is also similar to Case 1, but we need to account for the fact that the 3 letters that go in the third mailbox could be any 3 of the 4 letters. There are 4 ways to choose which 3 letters go in the third mailbox, and then there are 3 choices for the remaining letter, and 2 choices for where to put it. Therefore, there are 4 x 3 x 2 = 24 ways in which the letters could be placed in this case.

Case 4: The fourth mailbox has 3 letters.

This case is similar to Case 1, but we need to account for the fact that the 3 letters that go in the fourth mailbox could be any 3 of the 4 letters. There are 4 ways to choose which 3 letters go in the fourth mailbox, and then there are 3 choices for the remaining letter, and only 1 choice for where to put it. Therefore, there are 4 x 3 x 1 = 12 ways in which the letters could be placed in this case.

Therefore, the total number of ways in which one of the mailboxes could have exactly 3 letters is 27 + 24 + 24 + 12 = 87.

The total number of ways in which the 4 letters could be placed in the 4 mailboxes is 4^4 = 256.

Therefore, the probability that one of the mailboxes has exactly 3 letters is 87/256.

5, the equation just becomes (n+1)n!=720 just looking at this the answer is 5.  5!=120

It makes sense. but the system said its wrong.
I also tried 231 which worked?

100 integers in base 5

42 integers in base 7

Try 500. 6,500 is 13 times larger than 500

6,500 / 500 ==13 - which is what you want.

Let Xavier's three-digit number be represented as $ABC$, where $A$, $B$, and $C$ are digits representing the hundreds, tens, and ones digits, respectively. Then, Youssef's four-digit number is $3ABC$.

Since the four-digit number is 13 times as large as the original number, we have:

$$3ABC = 13 \cdot ABC$$

Simplifying this equation, we get:

$$3000 + 100A + 10B + C = 13(100A + 10B + C)$$

Expanding the right side, we get:

$$3000 + 100A + 10B + C = 1300A + 130B + 13C$$

Simplifying this equation further, we get:

$$1170A + 120B = 2999$$

Since $A$ and $B$ are digits, the left-hand side of this equation is an integer between 0 and 999. Therefore, the only possible value for $A$ is 2, and the only possible value for $B$ is 4. Substituting these values into the equation, we get:

$$1170(2) + 120(4) = 2520 + 480 = 2999$$

Therefore, Xavier's original number is $ABC = \boxed{240}$.

Ah sorry I meant 13 times larger. for some reason it didn't show

What does: u + v + w + x + y + z ==What?

Sorry, that is not the correct answer, thank you so much for giving it a try though!

 "If this four-digit number is  times as large as Xavier's original number, what number did Xavier write down"?  How many times larger?                             

i dont think it would be 196

stop using chatGPT to AI-generate answers.

Let the three-digit base 5 number be represented as $\overline{abc}_5$. Then, the value of this number in base 10 is $a\cdot 5^2+b\cdot 5^1+c\cdot 5^0$. We want this value to be expressed in base 7 with exactly two digits.

Let the two-digit base 7 number be represented as $\overline{de}_7$. Then, the value of this number in base 10 is $d\cdot 7^1+e\cdot 7^0$. We want this value to be equal to $a\cdot 5^2+b\cdot 5^1+c\cdot 5^0$.

Converting from base 5 to base 10, we have:

\begin{align*} \overline{abc}_5 &= a\cdot 5^2+b\cdot 5^1+c\cdot 5^0 \ &= 25a+5b+c. \end{align*}

Converting from base 7 to base 10, we have:

\begin{align*} \overline{de}_7 &= d\cdot 7^1+e\cdot 7^0 \ &= 7d+e. \end{align*}

Since we want these two values to be equal, we have:

\begin{align*} 25a+5b+c &= 7d+e \ 25a &= 7d+e-5b-c. \end{align*}

Since $0\le a,b,c\le 4$ and $0\le d,e\le 6$, we have:

\begin{align*} 0 &\le 7d+e-5b-c \le 6\cdot 7+6 = 48 \ 0 &\le 25a \le 4\cdot 25 = 100. \end{align*}

This means that $7d+e-5b-c$ can take on one of 49 possible values and $a$ can take on one of 4 possible values. Therefore, the total number of integers that can be expressed in base 5 using exactly three digits and expressed in base 7 using exactly two digits is $49\cdot 4 = \boxed{196}$.

We can break down the problem into two cases based on the value of x:

Case 1: x is 0 or 1.

If x is 0 or 1, then there are three possibilities for the value of y: 0, 1, or 2. For each of these values of y, we can break down the problem further based on the value of z:

If z is 0, then we need to find the number of solutions to the equation u+v+w = -x-y, where u, v, and w are nonnegative integers. This is a standard stars and bars problem, and the number of solutions is (3+2-1) choose 2 = 6 choose 2 = 15.

If z is 1, then we need to find the number of solutions to the equation u+v+w = -x-y+z-1. Again, this is a stars and bars problem, and the number of solutions is (3+2-1) choose 2 = 15.

If z is 2, then we need to find the number of solutions to the equation u+v+w = -x-y+z-2. Once again, this is a stars and bars problem, and the number of solutions is (3+2-1) choose 2 = 15.

Therefore, in this case, the total number of solutions is 3*(15+15+15) = 135.

Case 2: x is 2.

If x is 2, then y must be 0. We can break down the problem further based on the value of z:

If z is 0, then we need to find the number of solutions to the equation u+v+w = -x-y = -2, where u, v, and w are nonnegative integers. This is a stars and bars problem, and the number of solutions is (3+2-1) choose 2 = 6 choose 2 = 15.

If z is 1, then we need to find the number of solutions to the equation u+v+w = -x-y+z-1 = -1. This is also a stars and bars problem, and the number of solutions is (3+2-1) choose 2 = 15.

If z is 2, then we need to find the number of solutions to the equation u+v+w = -x-y+z-2 = 0. This is once again a stars and bars problem, and the number of solutions is (3+2-1) choose 2 = 15.

Therefore, in this case, the total number of solutions is 15+15+15 = 45.

Adding up the number of solutions from both cases, we get a total of 135+45 = 180 solutions to the equation.

Posted here: https://web2.0calc.com/questions/help-please_29046

Because there are three children, this is the same as saying that the three children can’t sit next to each other since the child in the middle would have another child on both sides. So, to answer this, we can calculate

(total number of arrangements) - (arrangements with all three children together):

Part 1: Total Number of Arrangements

This is just six people seated around a circle which has (6 - 1)! = 5! = 120 possibilities.

Part 2: Total Number of Arrangements with All Three Children Together

Treat the children as a single group which means we’re arranging four groups around the table so:

(total arrangements of 4 items in a circle) * (number of ways to arrange the 3 children)

(4 - 1)! * 3!

3! * 3!

6 * 6

36

This makes the total number of possible arrangements 120 - 36 = 84.

Given that x is a complex number with magnitude 2, we know that |x| = 2. We can also express x in polar form as x = 2(cos θ + i sin θ) for some θ.

Now consider the expression (3 + 4i)x^3. We can write this as:

(3 + 4i)x^3 = (3 + 4i)2^3(cos 3θ + i sin 3θ).

Similarly, x^5 can be expressed as:

x^5 = 2^5(cos 5θ + i sin 5θ).

The distance between (3 + 4i)x^3 and x^5 is given by the magnitude of their difference, which is:

|(3 + 4i)x^3 - x^5| = |(3 + 4i)2^3(cos 3θ + i sin 3θ) - 2^5(cos 5θ + i sin 5θ)|

= |8(cos 3θ + i sin 3θ) - 32(cos 5θ + i sin 5θ) + 4i|

= |8 cos 3θ - 32 cos 5θ + 8i sin 3θ - 32i sin 5θ|

= 8|(cos 3θ - 4 cos 5θ) + i(sin 3θ - 4 sin 5θ)|.

To find the maximum value of this expression, we want to maximize the expression inside the absolute value, which can be thought of as the distance between two points in the complex plane 

1

: (cos 3θ - 4 cos 5θ, sin 3θ - 4 sin 5θ) and (0,0). We need to find the angle θ that maximizes this distance.

This can be done by finding the angle between the two points, which is given by arctan((sin 3θ - 4 sin 5θ)/(cos 3θ - 4 cos 5θ)), and setting it equal to π. This is because the maximum distance between two points is achieved when the line connecting them is perpendicular to the x-axis, and hence has a slope of tan(π/2) = undefined.

Solving for θ yields:

θ = (1/2)(arctan(-4) + π/2)

Plugging this into our expression above gives an answer of 60.