Wiki Formatting Issues
\[\] Formatting Question #4: (https://brilliant.org/practice/perceptron-learning-algorithm/?chapter=perceptrons)
Which way of using bullets do you like better, left or right? (Compare the punctuation and capitalization.)
\[\] Formatting Question #3: (Poisson Distribution)
Which one do you like better, the one with or without colon at the end of the theorem title?
(Your answer will be applied to other cases such as sub-section titles and the like.)
Expected Value of Poisson Random Variable
Given a discrete random variable \(X\) that follows a Poisson distribution with parameter \(\lambda,\) the expected value of this variable is:
\[\text{E}[X]=\lambda.\]
Expected Value of Poisson Random Variable
Given a discrete random variable \(X\) that follows a Poisson distribution with parameter \(\lambda,\) the expected value of this variable is:\[\text{E}[X]=\lambda.\]
Expected Value of Poisson Random Variable:
Given a discrete random variable \(X\) that follows a Poisson distribution with parameter \(\lambda,\) the expected value of this variable is:
\[\text{E}[X]=\lambda.\]
\[\] Formatting Question #2: (Trailing Number of Zeros)
The way Andy Hayes puts a period/colon is different from the way the rest of us do. In the following example, please compare his original version and my edited version, and let him know about this if you agree with me.
Find the number of trailing zeros in \(452!.\) (Andy's)
For the formula above, \(n=452.\) Given that this factorial is in base ten, the goal is to find the highest power of \(5\) in \(452!.\) Therefore, \(p=5.\)
First, it is necessary to compute the number in base 5.
\[\begin{align} \left\lfloor \frac{452}{125} \right\rfloor &= 3 \\ 452 - 3 \cdot 125 &= 77 \\ \\ \left\lfloor \frac{77}{25} \right\rfloor &= 3 \\ 77 - 3 \cdot 25 &= 2 \\ \\ \left\lfloor \frac{2}{5} \right\rfloor &= 0 \\ \end{align}\]
Then \(452_{10}=3302_{5}.\)
The sum of the digits of 452 in base 5 is:
\[S_5(452)=3+3+0+2=8\]
Using the formula above, the highest power of 5 in \(452!\) is:
\[\begin{align} v_5(452) &= \frac{452-8}{5-1} \\ &= 111 \end{align}\]
Therefore, there are \(\boxed{111}\) trailing zeros in \(452!.\) \(_\square\)
Find the number of trailing zeros in \(452!.\) (Jimmy's)
For the formula above, \(n=452.\) Given that this factorial is in base ten, the goal is to find the highest power of \(5\) in \(452!.\) Therefore, \(p=5.\)
First, it is necessary to compute the number in base 5:
\[\begin{align} \left\lfloor \frac{452}{125} \right\rfloor &= 3 \\ 452 - 3 \cdot 125 &= 77 \\ \\ \left\lfloor \frac{77}{25} \right\rfloor &= 3 \\ 77 - 3 \cdot 25 &= 2 \\ \\ \left\lfloor \frac{2}{5} \right\rfloor &= 0. \end{align}\]
Then \(452_{10}=3302_{5}.\)
The sum of the digits of 452 in base 5 is
\[S_5(452)=3+3+0+2=8.\]
Using the formula above, the highest power of 5 in \(452!\) is
\[\begin{align} v_5(452) &= \frac{452-8}{5-1} \\\\ &= 111. \end{align}\]
Therefore, there are \(\boxed{111}\) trailing zeros in \(452!.\) \(_\square\)
\[\] Formatting Question #1: (Trailing Number of Zeros)
Starting from today, I want to number my questions about formatting so that I can use them later for an updated "Formatting Guideline." My first question is about boxing a number (or an expression). When the boxed number is the correct answer and is in the middle of the solution, that's fine. However, when it's right at the end of the solution, the box is followed by the small box \(_\square\) we use to end the solution, as shown in the \(2^\text{nd}\) and \(3^\text{rd}\) example problems below. If this happens, can I just go without the box around the number?
Give the prime factorization of \(10!,\) and find the number of trailing zeros of \(10!.\)
We have
\[\begin{array}{ccccccccccccccccccccc} 10! & = & 10 & \times & 9 & \times & 8 & \times & 7 & \times & 6 & \times & 5 & \times & 4 & \times & 3 & \times & 2 & \times & 1 \\ & = & 2^1 \times 5^1 & \times & 3^2 & \times & 2^3 & \times & 7^1 & \times & 2^1 \times 3^1 & \times & 5^1 & \times & 2^2 & \times & 3^1 & \times & 2^1, \end{array}\]
which can be rewritten as
\[10! = 2^8 \times 3^4 \times 5^2 \times 7^1.\]
The minimum power between \(2^8\) and \(5^2\) is 2. Therefore, \(10!\) has \(\boxed{2}\) trailing zeros. \(_\square\)
Find the number of trailing zeros in \(30!.\)
There are \(6\) multiples of 5 that are less than or equal to 30. Therefore, there are \(6\) numbers in the factorial product that contain a power of 5.
\(30!=30 \times 25 \times 20 \times 15 \times 10 \times 5 \times k\)
Note that one of these numbers, \(25,\) contributes a higher power of 5 to the product. \(25=5^2,\) and the other 5 multiples contain a \(5^1\) factor. Therefore, the number of trailing zeros of \(30!\) is \(\boxed{7}.\) \(_\square\)
Find the number of trailing zeros in \(500!\)
The number of multiples of 5 that are less than or equal to 500 is \(500 \div 5 =100.\)
Then, the number of multiples of 25 is \(500 \div 25 = 20.\)
Then, the number of multiples of 125 is \(500 \div 125 = 4.\)
The next power of 5 is 625, which is greater than 500.Therefore, the number of trailing zeros of \(500!\) is \(100+20+4=\boxed{124}.\)\(_\square\)