## python text is printed based on the divisibility of the number

The following code outputs a text based on the divisibility of the number. If the number is not divisible by 3.5 and 7 here, it will be printed out. Does using f string implicitly make the type box from integers to strings?

``````def convert(number):
result = ""
if (number % 3 == 0):
result += "Pling"
if (number % 5 == 0):
result += "Plang"
if (number % 7 == 0):
result += "Plong"
if(number % 7 != 0 and number % 5 != 0 and number%3 != 0):
result = f{'number'}
return result
``````

## Prove the divisibility test by \$ 7,11,13 for numbers with more than six digits

Prove the divisibility test $$7,11,13$$ for numbers with more than six digits

Attempt:

We know that $$7 cdot 11 cdot 13 = 1001$$, The for a six-digit number, for example, $$120,544$$Let's write it as
$$120544 = 120120 + 424 = 120 cdot1001 + 424$$
So we only check the divisibility of $$424$$ by $$7,11,13$$,

Know a number with more than six digits, for example: $$270060340$$.

$$270060340 = 270270270 – 209930$$
$$= 270 cdot (1001001) – 209930$$
$$= 270 cdot (1001000) + (270 – 209930) = 270 cdot (1001000) – 209660$$

So we check the divisibility of $$209660 = 209209 + 451$$, or only $$451$$,

The test says, however, that: z $$270060340$$we group three digits from the right:
$$270, 60, 340$$
then check the divisibility of $$340 + 270 – (60)$$,

How can you prove that?

## Divisibility of certain polynomials

Consider the finite sums
$$F_n (q) = sum_ {k = 1} ^ nq ^ { binom {k} 2}$$
with exponents the triangular numbers $$binom {k} 2$$, When $$n$$ strange, it seems that $$F_n (q)$$ does not factor over $$mathbb {Z} (q)$$, On the other hand, if $$n = 2 million$$ is just

QUESTION. is it true that $$F_ {2m} (q)$$ is divisible by the product
$$prod_ {j geq0} (1 + q ^ {m / 2 ^ j})$$
where the product lasts as long as $$m / 2 ^ j$$ is an integer.

Examples. Here is an example:
begin {align} (1 + q ^ 2) (1 + q) , vert & , F_4 (q); qquad (1 + q ^ 3) , , vert , F_6 (q); \ (1 + q ^ 4) (1 + q ^ 2) (1 + q) , , vert & , F_8 (q); qquad (1 + q ^ 6) (1 + q ^ 3) , , vert , F_ {12} (q). end

## Divisibility when a is odd.

Suppose that $$a | (4b + 5c)$$ and $$a | (2 b + 2 c).$$ Prove that if $$a$$ is so strange $$a | b$$ and $$a | c$$

So, since $$a | (2 b + 2 c)$$ that implies $$a | (4b + 4c)$$ in order to $$a | (4b + 5c)$$ After deducting two dividends, we receive directly $$a | c$$ but I can not understand why it should share $$b$$ if $$a$$ is odd.

## Probability – Divisibility of the square from the sum of two random variables

To let $$I_1, I_2$$ be two independent random variables.
$$I$$ is her sum, $$I = I_1 + I_2$$, Now consider the square of $$I$$.

$$I ^ 2 = (I_1 + I_2) ^ 2$$

We are interested in when $$I ^ 2$$ can be represented as the sum of two independent random variables. In other words, we can find two independent random variables $$X_1, X_2$$ so that
$$I ^ 2 = (I_1 + I_2) ^ 2 = X_1 + X_2$$

Suppose none of $$X_1, X_2$$ are constants.

In general, it's up NOT true.

Question 1: If $$I_1, I_2$$ are infinitely divisible, will be true above?

Question 2: If $$I_1, I_2$$ follows, what kind of special distribution then becomes true above?

## Infinite divisibility of log normals – MathOverflow

TL; DR: What is the low point of a piece of logarithmic normal distribution?

We know that logarithmic normals are infinitely divisible. What would be the low point of a root of lognormal?

More precisely, let us assume that $$X$$ is a log normal. Given an integer $$k> = 2$$, we now, since there is $$X_1, …, X_k$$ so that:

$$mathcal {L} (X) = mathcal {L} ( sum limits_ {i = 1} ^ {k} X_i)$$

From where $$mathcal {L}$$ denotes the distribution of a random variable.

$$Question:$$ Is there a way to simulate directly from $$X_i$$Law?

## Divisibility – What makes it so difficult to prove that \$ 3 \$ shares no odd perfect number?

To let $$sigma (x)$$ denote the Sum of divisors the positive integer $$x$$, The frequency Index from $$x$$ is then given by the formula $$I (x) = sigma (x) / x$$,

A number $$N in mathbb {N}$$ it is said that Perfect if $$sigma (N) = 2N$$, Equivalent, if $$N$$ is perfect $$I (N) = 2$$,

Euler showed that one odd perfect number $$M$$, if available, must necessarily have the form $$M = p ^ k m ^ 2$$, from where $$p$$ is the special prime number satisfying $$p equiv k equiv 1 pmod 4$$ and $$gcd (p, m) = 1$$,

Here is my question:

What makes it so difficult to prove that? $$3$$ does not share an odd perfect number?

MY ATTEMPT

So leave it now $$M = p ^ k m ^ 2$$ be an odd perfect number with a special prime $$p$$, Let's assume the opposite $$3 mid$$,

Since $$3 equiv 3 pmod 4$$, then $$p neq 3$$, (Actually, $$p$$ is prime with $$p equiv 1 pmod 4$$ implies that $$p geq 5$$).

So that means that if $$3 mid$$, then $$3 ^ 2 mid n ^ 2 mid$$, Accept that $$3 ^ 2 || M$$ (It means that $$3 ^ 2 mid$$ but $$3 ^ 4 nmid M$$). This implies that $$13 = 1 + 3 + 3 ^ 2 = sigma (3 ^ 2) mid sigma (M) = 2M$$, which means, that $$13 mid$$, This implies that
$$I (3 ^ 2) I (13) leq I (M),$$
provided that $$p = 13$$, or
$$1.60673 approx frac {9507} {5917} = I (3 ^ 2) I ({13} ^ 2) I ({61} ^ 2) I ({97} ^ 2) leq I (M) = 2$$
assumed $${13} ^ 2 || M$$ implies that
$$3 cdot {61} = 183 = 1 + 13 + {13} ^ 2 = sigma ({13 ^ 2}) mid sigma (M) = 2M$$
and
$$3 cdot {13} cdot {97} = 3783 = 1 + 61 + {61} ^ 2 = sigma ({61} ^ 2) mid sigma (M) = 2M,$$
where is there no contradiction.

postscript

I dare assume that an odd perfect number is not divisible everything the odd primes below $$100$$ (except for $$3$$. $$5$$, and $$7$$ – since we already know that $$105 nmid M$$). However, I have no proof. (Of course, I know that all it takes is a simple calculation.)

## Divisibility – When Bitcoin reaches \$ 100,000, how much is 0.01 Bitcoin worth in \$?

First of all, 0.01 bitcoin is 0.01 bitcoin, which is one-hundredth of a bitcoin.

So, if people were willing to pay \$ 100,000 for a Bitcoin-naive economy, the math would suggest they're willing to pay one-hundredth of that amount for a hundredth of a bitcoin, which is \$ 1,000.

## Group theory – divisibility of the group of invertible functions

Suppose we have the group $$G$$ of invertible functions over a particular set $$S subseteq mathbb {R}$$ under composition. I am interested in the divisibility of such a group. Take, for example $$S =[-1, 1]$$for each $$n in mathbb {N}$$you could try to find a function $$f in S$$ so that
$$underbrace {f circ f cdots circ f} _ {n text {times}} = sin ( frac { pi x} {2})$$
In general we want to satisfy $$forall n forall x exists y y ^ n = x$$, the usual axiom of divisibility. I suspect that these groups are not divisible, but it is known that they can always be embedded in a divisible group $$overline {G}$$, The construction of $$overline {G}$$ I've read that a series of wreath products and a direct limit are required. This usually makes things very abstract and can be difficult to identify $$f$$ as an element of $$overline {G}$$,

Is there a better way to visualize this construction? My goal is to understand the elements in $$overline {G}$$ as features over some larger (possibly not even real) $$S supseteq S$$,

## Number theory – divisibility properties of Pisano periods

To let $$(F_n)$$ the Fibonacci sequence and $$pi (m)$$ the pisano time of $$m$$ (ie the smallest period of $$F_n pmod {m}$$). There are many proven results over $$pi (m)$$, For example, this is known $$pi (p ^ a) mid p ^ {a-1} pi (p)$$, for any prime $$p$$ and an integer $$a geq 1$$, it is supposed that $$pi (p ^ a) = p ^ {a-1} pi (p)$$,

My question is a weak version of this assumption. For example, I want to prove that $$m ^ 2 mid pi (m)$$ for all positive integers $$m$$,

Since for $$m = p_1 ^ {a_1} cdots p_k ^ {a_k}$$it holds $$pi (m) = lcm ( pi (p_1 ^ {a_1}), lpoints, pi (p_k ^ {a_k}))$$Then it is enough to prove that
[
p^amid pi(p^{2a}),
]
For every prime $$p$$ and an integer $$a geq 1$$, There's the guess $$pi (p ^ a) = p ^ {2a-1} pi (p)$$We have some freedom, though $$a <2a-1$$ (ie. $$a> 1$$). Could someone help me?

Actually, even in the case of $$m mid pi (m ^ 3)$$ would be helpful. Thank you in advance!