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A couple well known rules for divisibility exist, let us start with the smaller one:

Divisibility by 3: A number is divisible by 3 if and only if it's digit sum is divisible by 3.

Divisibility by 5: A number is divisible by 5 if and only if it's last digit is either 0 or 5.

It seems what we need so far is some framework (at least be) able to

  1. pick out single digits and
  2. to calculate sums
  3. iterate on all resulting numbers $\mathbb Z \backslash \{0,1,\cdots,9\}$
  4. compare to a base case (set of integers $\subset \{0,1,\cdots,9\}$)

Multiplying with a vector $\bf v$ can achieve this. Summing the elements of a vector corresponds to scalar product with vector full of ones. Picking out last element corresponds to scalar product with vector $[0,\cdots,0,1]^T$.

Now to the tricky part. Can we mathematically prove that such a scalar product vector $v$ and set of end-up numbers $\mathcal S$ must exist for any divisor $ d\in \mathbb Z$?

Own work

  • Case $d=5, {\bf v}=[0,1]^T, \mathcal S = \{0,5\}$, for the subset $\{0,1,\cdots,99\}$
  • Case $d=7, {\bf v}=[3,1]^T, \mathcal S = \{7\}$, for the subset $\{0,1,\cdots,99\}$
  • Case $d=11, {\bf v}=[1,-1]^T, \mathcal S = \{0\}$, for the subset $\{0,1,\cdots,99\}$
mathreadler
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    I mean, we can take the sequence $10^k$ mod $d$, as we sometimes do. But I would assume that expecting the base cases to be in ${0,\cdots, 9}$ is too much to ask for large $d$. –  Nov 23 '19 at 11:08
  • Also, I don't understand what you do with $7$ exactly. Do you mean summing the unit digit to three times the integer quotient of $n$ by $10$? –  Nov 23 '19 at 11:14
  • for the number 21 we do scalar product $[3,1][2,1]^T = 3\cdot 2+1 \cdot 1 = 7 $ which is in our set so we have verified divisibility. If it was outside of the set it would have been disqualified. – mathreadler Nov 23 '19 at 11:16
  • Yeah. but what do you do with $946587$? –  Nov 23 '19 at 11:17
  • @Gae.S. : Then we need to make larger vector space and find longer vectors to do scalar products with. I think you are on to something with this sequence. – mathreadler Nov 23 '19 at 11:22
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    https://math.stackexchange.com/questions/3152587/largest-multiple-of-7-lower-than-some-78-digit-number/3152669#3152669 –  Nov 23 '19 at 13:16
  • @Gae raise it to a power, sufficient so it's length exceeds the order of 10 mod the prime. Base convert it to base p ? Convert it to a lower base, so it's length exceeds the order/cycle length in that base. Apply polynomial remainder theorem to change it to the same representation in a different base where you can tell figuring what changes). probably more. –  Nov 24 '19 at 00:28
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    you actually prompted me to think about the linked problem more, thank you. –  Nov 24 '19 at 16:53

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