So the problem is as follows:  Prove that if the sum of digits of a decimal $n$ is three's multiple, then n is three's multiple by direct proof. For example, $11234567$ is 3's multiple because $1+1+2+3+4+5+6+7=24$, and in fact, $11234567$ is $3744856 * 3$. Use $10^n \equiv 1 \pmod3$ if necessary.  I am having a lot of problems with this question seeing as it is a rather difficult problem. Could someone please help me solve this? Thank you!
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 $3\mid n$?
115364
Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.  By distinct it means that $(1, 0, 0)$ is a solution, but $(0, \pm 1, 0)$ counts as the same solution.  I can only seem to find two; namely $(1, 0, 0)$ and $( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3})$. Is there a method to finding a third or is it still just trial and error?
19148
Why is it that $\forall n \in N$, $2^n$ is not divisible by $3$?  I can prove it easily by induction, but I don't understand the intuition of why this is true. Could anyone please supply the intuition? [This has been answered.]  Moreover, I understand that there's also a proof of this fact that uses modular arithmetic. How does this proof go? [Also answered - it follows immediately from the definition of the modulus operation.]
8761
I'm working on a task where I'm a bit unsure if the answer I've got is correct.  Here is the task:     Show by induction that the following assertion is true for all natural   numbers $n$      $n^3 - n$ is divisible by $3$   Here is my answer:     For $n = 1$,   $n^3 - n = 1 - 1$   which is divisible by $3$      Assume the statement is true for some number $n$, that is, $n^3 - n$   is divisible by $3$. Now,      $(n + 1)^3 - (n + 1) = n^3 + 3n^2 + 3n + 1 - n - 1 = (n^3 - n) + 3(n^2 + n) = (n^3 - n) + 3n(n+1)$      which is $n^3 - n$ plus a multiple of $3$.      Since we assumed that $n^3 - n$ was a multiple of $3$, it follows that   $(n + 1)^3 - (n + 1)$ is also a multiple of $3$.      So, since the statement "$n^3 - n$ is divisible by $3$" is true for $n  = 1$, and its truth for $n$ implies its truth for $n + 1$, the statement is true for all whole number $n$.   I would appreciate if someone could go through the task and the answer and see if I've done this correctly.  Thanks a lot!
102472
Question     Numbers $n$ of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. For example:         $k=1$ then $n=10$.   $k=2$ then $n=31$.    $k=3$ then  $n=73$.   $k=4$ then $n=157.$      Conjecture:   the number $(2^k-1)\cdot 10^m+2^{k-1}-1$ where $m$ is the number of decimal digits of $2^{k-1}$ is never prime when it is of the form $7s+6$, that is when it is congruent to $6$ $\pmod 7$. Examples: $n=1023511$ ($k=10$)$\equiv 6 \pmod 7$ and thus it is composite $(1023511=19\times103\times523)$, $n=20471023$ ($k=11$) $\equiv 6 \pmod 7$ and thus it is composite ($20471023=479\times42737)$. With PFGW we arrived to $k=565000$ and all the $n's$ congruent to $6 \pmod 7$ are composite. According to Giovanni Resta's calculations in a post which has been canceled, there should be no probable prime congruent to 6 $\pmod 7$ upto k=800.000. The residue $6$ $\pmod 7$ occurs when either $m=6t+3$ and $k=3l+1$ or $m=6t+4$ and $k=3l+2$ with $k$ and $l$ some non-negative integers, but amazingly when it occurs the number is not prime. Can you find a counter-example or give a proof for the conjecture? Here a link to other interesting questions:  and  For primes of this form see:  The following vector contains all the exponents k&lt;=366800 leading to a prime  $[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$  Exponent $541456$ leads to another probable prime with residue 5 mod 7 and 325990 digits, but it need not be the next in increasing order.  Remark: we found five-in-a-row probable primes with res 5 mod 7. Probable primes with residue 5 are now twice frequent than expected. Exponents of these primes seem to be NOT random at all. Another thing I noticed, i don't know if it has some importance: the exponents leading to a probable prime $215, 69660, 92020, 541456$ are multiples of $43$. I noticed that $\frac{215}{41}, \frac{69660}{41}, \frac{92020}{41}, \frac{541456}{41}$ all have a periodic decimal expansion equal to $\overline{24390}=29^3+1$. This is equivalent to say that when k is a multiple of 43 and the number $10^{m}(2^{k}−1)+2^{k-1}−1$ is prime, then k is of the form $41s+r$ where r is a number in the set (1,10,16,18,37). Is there some mathematical reason for that?
35607
I was wondering why the last digit of $n^5$ is that of $n$? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks.
6479
For $n\in\mathbb{N}^*$ we note $s(n)$ the sum of the digits of $2^n$. For example $s(4)=7$ because $2^4=16$. Is it possible to find an integer $n$ such that $s(n)=s(n+1)$?
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.
36462
As the title implies.  I'm proving it by contradiction and I've already ruled out $p \equiv 0$ and $2$ (mod $4$).  I'm having trouble proving that it cannot be $\equiv 3$ (mod $4$). Namely, if $p \equiv 3$ (mod 4), then $m^2 + n^2 \equiv 1$ or $3$ (mod $4$). I know that $m^2 + n^2$ cannot be congruent to $3$ (mod $4$); but what about $1$?  Thank you.
80759
How do I solve this?  Prove that $3$ divides $14^{2n} -1$ for all $n = 0,\,1,\,2,\,3,\,\ldots$  My calculations are:     Let $n=1$ then $3|14^{2*1}-1 \implies 3|196-1 \implies 3|195$ which is true      Now $n = k$ and assume $3|14^{2k}-1$      let $n = k + 1$      $3|14^{2(k+1)}-1 \implies 3|14^2*14^{k+1}-1$      and $3|14^2-1$ so $3|14^2*14^{k+1}-1$      which makes $3 | 14^{2n}$ true.   I tried starting with $n = 0$ as it's the smallest base but I couldn't get it to work.
54300
Yeah so I'm kind of stuck on this problem, and I have two questions.   Is there a way to define a number mathematically so that it cannot be a multiple of $7$? I know $7k+1,\ 7k+2,\ 7k+3,\ \cdots$, but that will take ages to prove for each case.  Is this a proof? $$(7k + a)^3 \equiv (0\cdot k + a)^3 \equiv a^3 \bmod 7$$    Thank you.
82160
If we reverse the digits of $12$ we will get $21$. $12^{2}=144$. If we reverse its digits we will get $441$ which is $21^{2}$.  Here is the puzzle. How many such two digit numbers are there? Digits must be different.  We got an algebraic method to solve it if the square is a three digit number.It given here :  Let $a&gt;b$  $(10a+b)^{2}=100x+10y+z$ and $(10b+a)^{2}=100z+10y+x$  Difference between these two equations will lead to $a^{2}-b^{2}=x-z$  $x-z \leq 8$.Then  $a^{2}-b^{2} \leq 8$.  Since squares are of three digits $a &lt;4$.But since $a&gt;b ,a \neq 1$.  $b=1 \Longrightarrow a^{2} \leq 9$.Then $a=1,2,3$ but only possibilities are $2 $ and $3$.  $b=2 \Longrightarrow a^{2} \leq 12 \Longrightarrow a= 1,2,3$. But possibility is $a=3$. So next such pair will be $13,31$.  Higher values of $b$ will not work since $a$ cannot exceed 3. This is an explanation we obtained from a mathematics group.It works well if the square is a three digit number.What will happen if square is a 4 digit number... Or if we try to extend this problem for numbers having more than two digits ?
37565
This   states:      Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$   “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”      In other words, according to Example $1$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$   Should: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$ not be: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$ as everything in the left-hand side is cubed?
5350
I am pretty sure this question has something to do with the Least Common Multiple.    I was thinking that the proof was that every number either is or isn't a multiple of $3, 5$, and $8\left(3 + 5\right)$. If it isn't a multiple of $3,5$, or $8$, great. You have nothing to prove. But if it is divisible by one of them, I couldn't find a general proof that showed that it wouldn't be divisible by another one. Say $15$, it is divisible by $3$ and $5$, but not $8$.
79317
I have to show that this is divisible by both 2 and 3. For the next, part I have to show that it's divisible by 6 which from what I understand is the same as showing it's divisible by 2 and 3.
121952
My question is this : how many distinct two digit numbers can be produced from numbers $4, 3, 3, 1$? When applying the formula $$\frac{4!}{(4-2)!2!}$$ you come up with $6$, yet when doing the problem manually, I come up with $7$ numbers, namely $41, 43, 14, 13, 33, 31, 34$. Can anybody explain this discrepancy?  Thanks.
88424
Claim: A number is divisible by $4$ if and only if the number formed by the last two digits is divisible by $4$.   Here's where I've gotten so far.  Let $x$ be an $(n+1)$-digit number. So $x= a_na_{n-1} \dots a_2a_1a_0$. If $a_1 = 0$ and $a_0 =0$, then $x$ is a multiple of $100$ and therefore clearly divisible by $4$. So we must deal with the case when $(a_1 \neq 0 \lor a_0 \neq 0)$.  Then if $10a_1 + a_0 \equiv 0 \mod 4$ is true, then $x$ is divisible by $4$.  Do I need to do anything else or is this done? I feel like it's not quite complete, but I'm not sure how to proceed.