How can I prove that $9|n$ if and only if $9$ divides the sum of the digits of $n$, where $n \in \mathbb{N}$
122068
Show that the number 9 divides the number $ m$ if and only if the sum of the digits of the number $ m $ is divisible by 9.  Show that the number 3 divides the number $ m$ if and only if the sum of the digits of the number $ m $ is divisible by 3.  Are semigroup and monoid theory could be useful?
63842
Let $n\in\mathbb N$ and let $a_1,\ldots,a_n$ be natural numbers smaller than $10^n$. Write each $a_k$ in base $10$ and add $0$'s to the left of each decimal expansion, if needed, so that each $a_k$ is written with $n$ digits. Consider the matrix such that the entries of the $k$th row are the digits of $a_k$ (written in the same order). Let $d\in\mathbb N$ be such that $d$ divides each $a_k$. Prove that $d\mid\det A$.   For instance, suppose that $n=4$ and that your numbers are $3876$, $2784$, $684$, and $8388$, each of which is a multiple of $12$. Then$$A=\begin{bmatrix}3 &amp; 8 &amp; 7 &amp; 6 \\  2 &amp; 7 &amp; 8 &amp; 4 \\ 0 &amp; 6 &amp; 8 &amp; 4 \\  8 &amp; 3 &amp; 8 &amp; 8\end{bmatrix}$$and $\det A=-360$, which is, in fact, a multiple of $12$.  I learned about this problem yesterday. I found it quite cute and I decided to share it with all of you.  Note: I know how to prove it.
60821
Let $M$ be the set of natural numbers that can be written using only $0$ and $1$(in the decimal system). Prove that, for any natural number $k$ $\exists m \in M $ s.t   i) $m$ has exactly $k$ $1$'s, and ii) $m$ is divisible by $k$. (e.g, if $k=3$, then the number $101010 \in M$ is divisible by $3$)    Any hint or solution is welcome. If anyone have any hint in mind please give me step by step. Problem seems to be very interesting to me.
72318
I have not been able to solve this problem. Any insights would be appreciated!   Let $x, n > 1$ be integers. Suppose that for each $k > 1$ there exists an integer $a_{k}$ such that $x − a_k^n$ is divisible by $k$. Prove that $x = A^{n}$ for some integer $A$.
101017
I want to prove that:  $2^{n-1}\mid n!$ if and only if $n$ is a power of $2$.
24673
The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three.  This is quite helpful for determining if really large numbers are multiples of three, because we can recursively apply this rule:  $$1212582439 \rightarrow 37 \rightarrow 10\rightarrow 1 \implies 3\not\mid 1212582439$$ $$124524 \rightarrow 18 \rightarrow 9 \implies 3\mid 124524$$  This works for as many numbers as I've tried.  However, I'm not sure how this may be proven.  Thus, my question is:     Given a positive integer $n$ and that $3\mid\text{(the sum of the digits of $n$)}$, how may we prove that $3\mid n$?
19737
I'm trying to teach myself some number theory. In my textbook, this proof is given:  Example (2.3.1) Show that an integer is divisible by 3 if and only if the sum of its digits is a multiple of 3. Let $n=a_0a_1\ldots a_k$ be the decimal representation of an integer $n$. Thus $n=a_k+a_{k-1}10+a_{k-2}10^2+\cdots+a_010^k$ where $0\le a_i&lt;10$ The key observation is that $10\equiv1\pmod3$, i.e., $[10]=[1]$. Hence $[10^i]=[10]^i=[1]$, i.e., $10^i\equiv1\pmod 3$. The assertion is an immediate consequence of this congruence.  I don't understand the last statement. Why does it follow from that congruence?
7814
If "x" is a 9 digit number , which contains digits from 1 to 9 which ends in 5 . Prove that it can't be a perfect square ( digits are not to be repeated).  Please suggest the solution of this question.
7156
Prove that   $$\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$$   I can't find counting interpretations for either of the sides. A hint of "if $S$ is a subset of $\{1, . . . , n\}$ and $S^\prime$ is its complement then $|S| + |S^\prime| = n$" was also given, but I still don't know how to begin.
72410
Now I have proved the following two problems:      (1) Prove that a number and the sum of its digits have the same remainder upon division by 9.    (2) Given an integer, consider the difference between the sum of the digits in odd positions (counting from the right) and those in even positions. Show that the difference has the same remainder as the number itself when divided by 11.    Then, I need help on the following problem:      Partition the digits of a number into groups of two and three, respectively, starting from the right. Treat these as two- and three-digit numbers, respectively, and proceed similarly as in the two problems above. What divisibility rules arise?    In fact, I did not quite get what this problem asks for and not clear about this process. Could anyone give me some hints on this, please?  Thanks in advance.
1706
I am having trouble with the following proof:     For every positive integer $n$: $$1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$$   My work: I have tried to add $\frac{1}{(k+1)^2}$ to $2-\frac{1}{k}$ in the inductive step and reduce it down to $2-\frac{1}{k+1}$ but cannot do so. I am beginning to think that my entire approach is wrong.
120789
Prove that if $n \in \mathbb{N}$, then $$\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2$$ where $d(n)$ is the divisor function. I have know only information is $d(n)=\sum_{d|n}{1}$.
69792
I'm solving a problem which asks me to find how many 5-digit numbers with distinct digits are there which are also divisible by $3$. I am familiar that there are $9\cdot9\cdot8\cdot7\cdot6=27216$ five-digit numbers with distinct digits. I know that a natural number $n$ is divisible by $3$ iff the digits sum up to a number divisible by $3$. Now, brute-forcing by a computer algorithm, i found that the result is $20928$. However, I am interested, if there is a mathematical solution for this. I know how to find amount of numbers divisible by $4$ or $5$. We just fix some amount on the last two (or last one) digit. But criterion for divisibility by $3$ is very different from those and I have no idea how to start. Eventually, I would like to find a general formula for $n$-digit numbers.  Edit: Equivallently we want to find how many solution does the equation $a+b+c+d+e=3k$ for $a,b,c,d,e\in\{0,1,2,...,9\}$, $a\neq 0$ also $a,b,c,d,e$ are all distinct and $k\in\mathbb{Z}$ (the highest possible $k$ is in this case $k=11$, the smallest being $k=5$)  Edit (2): According to the ideas from the answer. Make $A=\{0,3,6,9\},B=\{1,4,7\}, C=\{2,5,8\}$. Let's go through the cases that in the number $n$ there are $0,1,2,3,4$ digits from $A$.  i) Suppose none of $A$ is in $n$. We can't form a number divisible by three. We have to take 5 numbers, that is $3$ from $A$ and $2$ from $B$ or $2$ from $A$ and $3$ from $B$. None of the expressions $3(3k+1)+2(3k+2)$ and $2(3k+1)+3(3k+2)$ are divisible by three.  ii) Let $1$ digit from $A$ be in $n$. The amount of ways to choose the digit is $5$. Now we are left to choose $4$ digits. The only possible way is to choose $2$ from $A$ and $2$ from $B$. The amount of ways to do this is ${3\choose 2}^2=9$ which together gives $5\times 9$ ways which can be permuted and alltogether $5\times 9\times 5!=5400$ possible ways.  iii) Let $2$ digits from $A$ be in $n$. The amount of ways to choose the digit is ${5\choose 2}=10$. We are left with $3$ digits. Either all of them are from $A$ or all of them are from $B$. That's $2$ possible ways and all together $20$ possible ways which we permute to obtain $20\times 5!=2400$ possible ways.  iv) Let $3$ digits from $A$ be in $n$. The amount of ways to choose the digits from $A$ is ${5\choose 3}=10$. We are to choose $2$ digits. We must take exactly one from $B$ and exactly one from $C$. The amount of ways to do this is ${3\choose 1}^2=9$. Alltogether $9\times 10$ possible ways which we permute to get $90\cdot5!=10800$ possible ways  v) In the last case where all four digits from $A$ are present, there is no way how to make the number divisible by $3$.  Summing this all up gives $5400+2400+10800=18600$ which is less than what my algorithm yielded. We also have to subtract these which begin with $0$. Can anyone spot the mistake? (If i did any?)
119953
I am having trouble proving the two following questions:    If $p|N$, $q|N$ and gcd(p,q)=1, then prove that $pq|N$ If $x$ is non zero positive integer number, then prove that $3|x$ if and only if 3 divides the sum of all digits of $x$.   For both questions I tried to use theorems of discrete mathematics, but I could not find the way to solve them.
82857
I accidentally found that $9^n\equiv 9\pmod{24} \quad \forall n \in \Bbb N^*$ But don't know how to prove it in an exact way. Please help
14435
I have tried to sopve this question and found the answer, but i cant find a proof. My answer is $1,3,9,27$. Thanks in advance to those who help me.
44723
Let the sum of all integers $n$ such that $(2n^2+9/n+3)$ is an integer be $A$, what is $|A|$?  I tried expressing it to:  $k$ is an integer, $(n+3)k + 0 = 2n^2 + 9$, $n(2n-k) + 3(3-k) = 0$, from which I couldn't move on,  I also tried $(n+3)k + 0 = 2n^2 + 9$, $k = 2n$, $2n^2 - 2n^2  + 9 - 6n$. $9-6n/n+3$ is an integer, $9 - 6n = 3(3-n)$ though now I also do not know how to move on.