I still remember the feeling, when I learned that a number is divisible by $3$, if the digit sum is divisible by $3$. The general way to get these rules for the regular decimal system is asked/answered here: . Now I wonder, what divisibility rules an alien with $12$ (or $42$) fingers would come up with?  So let $n=\sum_k c_k b^k$ be the representation with base $b$. Looking at some examples shows indicate that $n$ is divisible by $b-1$, if $\sum_k c_k$ is. This seems to be a poor man's extension to the decimal divisibility by $9$ rule.  The answer to the above mentioned question, says that "One needn't memorize motley exotic divisibility tests. ". Motley exotic divisibility tests are very welcome here!
122467
How could we make the same division rules for other number systems, like in our decimal system: a number is divisible with 2 if it's last digit is 0,2,4,6,8, by 3 if the sum of digits is divisible with 3, and so on.  Can we do the same rules for g-based number systems too, where g is not 10, and greater than 1?
32627
Suppose we have a natural number $N$ with decimal representation $A_kA_{k-1}\ldots A_0$. How do I prove that if the $\sum\limits_{i=0}^kA_i$ is divisible by $9$ then $N$ is divisible by $9$ too?
13240
Is there a direct way to determine how many digits a power of 2 will contain without actually performing the multiplication?  An estimation would help as well if there is no absolute solution.  EDIT: In both decimal and binary bases.
115516
i have looked at/practiced several methods for solving ex: $7^{9999}$. i have looked at techniques using a)modulas/congruence b) binomial theorem c) totient/congruence d) cyclicity.   my actual desire would be a start to finish approach using totient/congruence. i have figured out how to work the individual steps but not how to combine them and in what order to be able to solve any "end digits" questions.
93127
If the three digit numbers: $x17, 3y6, 12z$ where $x,y,z$ are integers from $0-9$ are divisible by a fixed constant $k,$ then the determinant   $$\left|\begin{matrix} x &amp; 3 &amp; 1 \\ 7 &amp; 6 &amp; z \\ 1 &amp; y &amp; 2\end{matrix}\right|$$  must be divisible by?
107067
How do you prove that a number is divisible by 3 iff the sum of its digits is divisible by 3? Please avoid using the modulus or any other operations or advanced terminology from number theory because I have never taken that class.
37387
How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8$ and $9$ which are divisible by $3$ and $5$, without any of the digits repeating? For the number to be divisible by $5$ it must end with $0$ (or $5$, but we don't have it in the given problem). So we have $V_6^4$ numbers which are divisible by $5$ but how can we exclude those which aren't divisible by $3$?   $120$
118577
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$ Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.  How to calculate the missing digits? I know that large factorials can be estimated using : $$15! \approx \sqrt{2\pi\cdot 15} \cdot \left(\frac{15}{e}\right)^{15}$$ which is not feasible to calculate by hand. The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11: $1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 \mod \phantom{1}9 \equiv \,34 + d_0 + d_1 \mod \phantom{1}9 \equiv 0 $ $-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 \mod 11 \equiv d_0 + d_1 \mod 11 \equiv 0 $ The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
61148
Here is the proof of the converse:  Iff a number $n$ is divisible by $3$, then the sum of its digits is also divisible by $3$.  Proof:  We know $n \mod 3 = 0$. By the basis representation theorem, $n$ can be re-written as $n_k10^k + n_{k-1}10^{k-1} + \cdots + 10n_1 + n_0 \equiv 0 \mod 3$. By the modular arithmetic addition rule, we can take $n_k10^k \mod 3 + n_{k-1}10^{k-1} \mod 3 + \cdots + 10n_1 \mod 3 + n_0 \mod 3$. But then we just get $n_k + n_{k-1} + \cdots + n_1 + n_0$, since $10 \mod 3 \equiv 1$.   Now suppose that the sum of the digits was divisible by 3. How can I prove that the representation in base $10$ is divisible by $3$?
7880
A number is divisible by $2$ if it ends in $0,2,4,6,8$. It is divisible by $3$ if sum of ciphers is divisible by $3$. It is divisible by $5$ if it ends $0$ or $5$. These are simple criteria for divisibility.  I am interested in simple criteria for divisibility by $7,11,13,17,19$.
122378
I stumbled upon the following in a book:     till 400 all even numbers will be divisible by 2 ( 200 even numbers)   remaining 200 odd numbers  1 3 5 7 9 ..... 399 200/3 = 67 will be   divisible by 3, as we have discarded all even numbers and out of   remaining numbers 2nd number is 3 then 1 + 198/3 = 67 numbers are   divisible by 3      now we have discarded all multiples of 2 and 3 remaining numbers will   be  200 - 67 = 133;   1 5 7 11 13 ... so on out of these 133 numbers 2nd   number is 5 so numbers divisible by 5 will be 1 + 131/5 = 1+26 = 27      now remaining numbers = 133 - 27 = 106; 1 7 11 13 .. so on again 2nd   number is 7 , so numbers divisible by 7 = 1 + 104/7 = 1+14 = 15      total numbers from 1 - 400 divisible by any of 2 3 5 7 = 200+67+27+15   = 309   But I can't seem to understand how the following is derived? :     as we have discarded all even numbers and out of remaining numbers 2nd   number is 3 then 1 + 198/3 = 67 numbers are divisible by 3   How does the result come out to be "1 + 198/3" ?
15233
We know that $1331$ is divisible by $11$. As per the $11$ divisibility test, we can say $1331$ is divisible by $11$. However we cannot get any quotient. If we subtract each unit digit in the following way, we can see the quotient when $1331$ is divided by $11$.  $1331 \implies 133 -1= 132$ $132 \implies 13 - 2= 11$ $11 \implies 1-1 = 0$, which is divisible by $11$.  Also the quotient is, arrange, all the subtracted  unit digits (in bold and italic) from bottom to top, we get $121$. Which is quotient. I want to know how this method is working? Please write a proof.
102849
While solving the question of  a question arrived in my mind; the question was  Everyone knows the divisibility rule of $13$.  Test for divisibility by $13$: Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number.   For Example : $$50661$$ $$5066+4=5070$$ $$507+0=507$$ $$50+28=78$$   and $78$ is $6\times13$, so $50661$ is divisible by $13.$  Please help me!!!
32197
For what values of $m,n$ natural, do $2^n - 1$ is divisible by $2^{m+n} - 3^m$?  Thank you very much.
56919
prove if we want that the sum of some fractions be $1$ and the denominators of one of them is $d$ then another denominators should divisible by $d$ or $d$ should be divisible to another denominators.  It seems to be easy I tried to prove it.I first tried some cases.  $1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$  Here we can see that $6$ is divisible by $3$. Also here $6$ is divisible by $2$ .But I want to prove one of the denominators but here two of them is possible.After trying a lot I cannot found any proofs.Any hints?  update1: the numerator should be prime
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.
92025
I am trying to calculate the last two digits of $3^{400}$, but working $\bmod {100}$ is proving intractable. I did the last digit case pretty handily, and now I'm stuck at last two.