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Divisibility rules of 1,2,3,4,5,6,8,9 are first or second grade math. Are there any divisibility rules for numbers with factors including 7. I noticed that the digits of 7x starting with x=1 to x=5 have digits that add up to 7, 5, 3, 10, and 8.

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marked as duplicate by Zev Chonoles, Rory Daulton, Micah, user2566092, Joel Reyes Noche Jul 3 '15 at 0:39

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  • $\begingroup$ I would add $10$ and $11$ to that list. Those aren't difficult. $8$ is actually trickier in my opinion. $\endgroup$ – Arthur Jul 2 '15 at 22:33
  • $\begingroup$ The Universal Divisibility Test is quite simple (and memorable!) $\ \ $ $\endgroup$ – Bill Dubuque Jul 3 '15 at 1:27
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Test #1. Take the digits of the number in reverse order, from right to left, multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary. Add the products. This sum has the same remainder mod 7 as the original number!

Example: Is 1603 divisible by seven? $3(1)+0(3)+6(2)+1(6)=21$ is divisible by $7$, so $1603$ is.

Test #2. Remove the last digit, double it, subtract it from the truncated original number and continue doing this until only one digit remains. If this is 0 or 7, then the original number is divisible by 7.

Example: $1603$; $160-2(3)=154$; $15-2(4)=7$, so 1603 is divisible by 7.

Source: https://www.math.hmc.edu/funfacts/ffiles/10005.5.shtml

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If you have a $n$ digit number, take the leading $n-1$ digits and treat this as a number $M$ (i.e., set the last digit of your original number to $0$ and then divide by $10$), and let the last digit of your original number be $N$, and then your number is divisible by $7$ if and only if $M - 2N$ is divisible by $7$. So you can keep reducing the number of digits using this rule until you get your answer. Try with some numbers like 63 and 91.

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  • $\begingroup$ I started writing this answer before I saw the duplicate question comment. So I am also going to vote to close. $\endgroup$ – user2566092 Jul 2 '15 at 22:40
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Split the decimal number into digits and re-compose them, but multiply them by powers of $3$ instead of $10$. If the resulting number is too big, repeat the routine until you know the answer – all intermediate results are congruent modulo $7$.

Example:
$$\begin{align} 704 \to && \ 7\times 3^2 + 0\times 3^1 + 4\times 3^0 & = & 67 \\ 67 \to && \ 6\times3^1 + 7\times 3^0 & = & 25 \\ 25 \to && \ 2\times 3^1 + 5\times 3^0 & = & 11 \\ 11 \to && \ 1\times 3^1 + 1\times 3^0 & = & \color{red}4 \end{align}$$

The sequence $1, 3, 2, 6, 4, 5$ mentioned by @Phugo are the remainders of $3^i$ when divided by 7, so they serve the same way as powers of $3$, giving the same result with faster convergence.

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