# Tag Info

29

One way is the total number of leaves of a (single) rooted tree in which each leaf is minimally linked to the root by exactly $n-1$ edges, and which has the following property: the root has $2$ children, each child of the root has $3$ children, each child of each child of the root has $4$ children, and so on until the leaves are reached. A natural term for ...

13

HINT: The number is a multiple of $9$ and a multiple of $11$.

10

If $m$ is not a prime, it factors into two numbers smaller than $m-1$, thus we have $m \mid (m-1)!$ and $(m-1)! \equiv 0 \mod m$ if those factors are different. If these numbers are not different, then it is sufficient to have $\sqrt{m}$ and $2\sqrt{m}$ as factors. As we have $2\sqrt{m}<m-1$ for $m\geq9$, the only other case is $m=4$, in which case we ...

9

Hint: $n! = n \cdot (n - 1)!$.

7

Look closer at your graph! (I am going to use $k$ and $n$ here.) I get e.g. $$\frac{1}{100000}\left(\sum_{k=1}^{100000}\ln(k)\right)\approx \ln(100000)-0.999933$$ This suggests an asymptotic of $\ln(n)-1$. Indeed: $$\begin{array}{ll} \displaystyle \frac{1}{n}\sum_{k=1}^n\ln(k) & \displaystyle =\ln(n)+\frac{1}{n}\sum_{k=1}^n \big(\ln(k)-\ln(n)\big) \\ ... 7 We will use the following facts: (i) The extension, to a larger domain, of a non-elementary function is also non-elementary; (ii) The derivative of an elementary function is also elementary; (iii) The product of finitely many elementary functions is also elementary; (iiii) The product of an elementary function times a non-elementary ... 5 Using the fact that n!\sim_{\infty }\sqrt{2\pi n}\left(\frac{n}{e}\right)^n,$$\lim_{n\to\infty }\frac{n!}{\sqrt{3n}\left(\frac{n}{2}\right)^n}=\lim_{n\to\infty }\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{\sqrt{3n}\left(\frac{n}{2}\right)^n}=\lim_{n\to\infty }\sqrt{\frac{2\pi}{3}}\underbrace{\left(\frac{2}{e}\right)^n}_{\to 0}=0$$and thus ... 5 Here's a geometric visualization in higher dimensions. You can take a hyper-cube in dimension d (basically the Cartesian product of n copies of the interval [0,c] for any c > 0 that you want), and then you triangulate (i.e. partition) into equal volume simplices (a simplex in d dimensions is a full dimensional convex hulls of d+1 points, i.e. ... 4 The way I see n! is a hybrid of avid19's and Dave L. Renfro's visualizations: I imagine n people lining up one by one. I think it really helps to imagine people or animals or fruits or something, rather than boring symbols: that's the way it's done in Burns and Weston's Math For Smarty Pants, and it seems to have made quite an impression on me. My ... 4 You know that 34! is divisible by 9, so, because the sum of all other digits is 141, you know that c+d=3 or c+d=12. Now do the same for divisibility by 11 (remember the alternating sums criterion?). 4 A proof I found a while ago entirely relies on creative telescoping. Since \frac{1}{n^2}-\frac{1}{n(n+1)}=\frac{1}{n^2(n+1)},$$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq ...

4

Notice, $$9!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9$$$$=362880=2^7\times 3^4\times5\times 7$$ total number of odd factors $$=1\times5\times 2\times 2=40$$ But, the odd factors of form $(3m+2)$ are only two which can be found by putting $m=1$ & $m=11$ which are $3\times 1+2=5$ & $3\times 11+2=35$ Hence the sum of the odd ...

3

The possible numbers are $5,11,17,23,29,35$. So, we have at least $5+35=40$. From the given options, the answer is $(A)$.

3

$9!$ = $1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9$. Of all the odd factors, only $5$ is of the form $3m + 2$. We can have $5$, and the only possible number to combine it is $7$ to give $35$ which is also of the form $3m + 2$. So these are the only 2 factors and their sum is $40$

3

If $n!>n$, then $$(n+1)! = (n+1)n! > (n+1)n > n+1$$ as long as $n>1$. Now, $3!>3$, and by induction we have proved that $n!>n$ for all $n \geq 3$. Since $1!=1$ and $2!=2$, the assertion is proved.

3

Obviously, $n\ne0$, and we can simplify: $$(n-1)!=1.$$ By inspection, $n-1=0$ and $n-1=1$ are solutions. All other factorials have a factor that makes them exceed $1$.

2

This might not be what you're looking for, but I visualize a factorial as a process. $5!$ is how many ways you can arrange 5 things. I visualize arranging 5 things. Not a representation as dots but personally it's powerful.

2

We are given that n!= \left\{\begin{align}1\quad \text{for}\quad n=0\\n\cdot(n-1)!\quad \text{for}\quad n>0\end{align}\right. and wish to show that $$n! = \prod_{i=1}^ni$$ We start with $n=1 \implies 1! = 1\cdot(0!) = 1 = \prod_{i=1}^1 i$ Now, let's assume $(n-1)! = \prod_{i=1}^{n-1} i$, then $$n! = n\cdot(n-1)! = n\prod_{i=1}^{n-1} i = ... 2 Stirling's formula says the ratio of$$ \sqrt{2\pi x\,{}} \, \frac{x^x}{e^x} $$to x! approaches 1 as x\to\infty. You seem to have a crude approximation of Stirling's formula. (I think Stirling's contribution to this may have been the value of the constant factor. de Moivre earlier showed that the ratio of \sqrt{x}\ \dfrac{x^x}{e^x} to x! ... 2 This provides another way. Let a_{n} > 0 for all n \geq 1. Then it can be shown that$$\liminf a_{n+1}/a_{n} \leq \liminf (a_{n})^{1/n} \leq \limsup (a_{n})^{1/n} \leq \limsup a_{n+1}/a_{n}.$$Thus if there is some l \in \mathbb{R} such that a_{n+1}/a_{n} \to l then (a_{n})^{1/n} \to l. Let a_{n} := n^{n}/n! for all n \geq 1. Then ... 2 n! = n(n-1)! So, start simplifying the first part of your equality. \frac{n!}{(n-1)!} = \frac{n(n-1)!}{(n-1)!} = n. Do you see how to continue? 1 The correct constants are twice the value of \pi=3.1415\ldots and e=2.718281828459045\ldots, not 3 and 2. That's why it might seem close, that is,$$n!\sim n^n e^{-n} \sqrt{2\pi n}$$1 I like Knuth's notation for falling factorials better:$$ \alpha^{\underline{h}} = \alpha \cdot (\alpha - 1) \cdot \dotsm \cdot (\alpha - h + 1) $$First note that:$$ \Delta \alpha^{\underline{h}} = (\alpha + 1)^{\underline{h}} -\alpha^{\underline{h}} = (\alpha + 1) \cdot \alpha^{\underline{h - 1}} - \alpha^{\underline{h - 1}} \cdot (\alpha ...

1

They are not equal; ${n \choose 2} = \frac{n(n-1)}{2}$. But the difference between them is $n/2$, which is small enough that it can be rolled into the correction term in your link.

1

Here, the factorials don't really matter because $2!=2$, but in general, we do need to take factorials. Consider a simpler problem: How many anagrams does the word "AAAB" have? The answer is clearly $4$, since once we know where the "B" is, we know the rest of the letters are "A". To get there, we have to consider that there are $4!=24$ ways to arrange $4$ ...

1

The factors that you are looking for are not divisible by 2 (because they are odd) and not divisible by 3 (because they are divisible by 3). We have 9! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9. If we remove the factors 2 and 3, then all that is left is 5 x 7, and the only factors that are odd and not divisible by 3 are 1, 5, 7 and 35. Of these, only 5 and 7 ...

1

If I properly read the question, you are concerned by $$a_n=\frac{(n!)^{\frac{1}{n}}}{n}$$ So $$\log(a_n)=\frac 1n \log(n!)-\log(n)$$ As usual when working with factorials, Stirling approximation is useful. You will in particular find this very nice inequality $$\sqrt{2\pi}n^{n+\frac 12}e^{-n} \leq n! \leq en^{n+\frac 12}e^{-n}$$ which would give you good ...

1

To prove that $n>2\implies n!>n$ you can use induction. Base case: $3!=6>3$ Induction step: if $n!>n$ and $n>2$ then $(n+1)!=n!(n+1)>n(n+1)>n+1$.

1

Stirling's formula unfortunately has potentially big errors - the approximation does not mean that $$n! -\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n\to 0,$$ only that $$\frac{n!}{\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n}\to 1.$$ So an approximation of $n!$ is not of much use. Basically, the above formula is "trivial" one's you know Wilson's theorem: ...

1

$$n!-(n-1)!=n(n-1)!-(n-1)!=(n-1)!\left( n-1 \right)$$

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