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Find the amount of both positive and negative multiples of $7$ in the range of $(-4,000,000, 4,000,000)$ are there such that the only numbers that are used as digits are $4, 2,$ and $0.$ For example, one such number is $42=6 \cdot 7.$

I have not gotten much progress, but so far I realized that for the millions digit we can only have $2$ or $0.$ For all other digits we can have $4, 2,$ or $0.$ This gives $2 \cdot 3^6=1458$ possible numbers in the range of $[0, 4,000,000).$ We can later double the amount of numbers in this range then subtract $2$ (since we don't count $0$) for our desired outcome. However, I'm not sure how to compute how many of these $1458$ possible numbers are multiples of $7$ other than a large, messy casework bash. Can I have some help?

Edit: I still have not gotten much progress. I do understand how we may divide by 7 to get an estimation of $208$ for half the interval, but I am wondering if there is a way to rigorously show that this is indeed the correct amount of possible numbers instead of an estimation. Thanks in advance.

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  • $\begingroup$ You might guess without much justification that it could be about $1458/7 \approx 208.29$ (for the positive cases) and indeed it is $\endgroup$
    – Henry
    Commented Sep 5, 2021 at 22:23
  • $\begingroup$ @Henry How would we guess that around a seventh of the numbers in that form are a multiple of $7?$ Also, is there a more rigorous way to compute it other than approximation? $\endgroup$
    – mathisfun
    Commented Sep 5, 2021 at 22:24
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    $\begingroup$ The guess is simply that about a seventh of positive integers up to a limit are divisible by $7$ and so about a seventh of of a particular subset might also be. But this may be wrong either because the subset is extra special (suppose we were restricted to digits $0$ and $7$) or because this is only an approximation $\endgroup$
    – Henry
    Commented Sep 5, 2021 at 22:31
  • $\begingroup$ Oh I see. It seems that for our case the guess results with the correct amount. Do you think there a more rigorous, but not very messy strategy to doing this? $\endgroup$
    – mathisfun
    Commented Sep 5, 2021 at 22:35
  • $\begingroup$ Ask the same question for a modified base 3 $\endgroup$ Commented Sep 6, 2021 at 15:23

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First we find the greatest $7$-multiple under $8$ million. The we factor it's cofactor of $7$. Then we express the result as sums, products, or differences of powers of $2$. $$ 4,000,000-(-4,000,000)\\ =8,000,000\\ =1+7,999,999\\ =1+7(1,142,857)\\ =1+7({199\cdot 5743}) $$

\begin{align*} 1&=2^0\\ 7&=4+2+1\\ &=4+2+2^0\\ 199&=128\quad\space\space+64\quad\space+7\\ &=2^{(4+2+2^0)}+2^{(4+2)}+(4+2+2^0) \end{align*}

\begin{align*} 5743&=4096\quad+1024\quad+512\quad+128\quad-16\quad-1 \\&=2^{(2(4)+4)}+2^{(2(4)+2)}+2^{(2(4)+1)} +2^{(2(4)+4)}-2^{2(4)}-2^0 \end{align*}

If we now "write" the [binary] expression for the product of $199$ and $5743$ we should have the desired result..

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  • $\begingroup$ Is there a process on how you factored $1,142,857?$ Also is there motivation behind expressing the results as sums, products, or diferences of power of $2?$ Also, I'm a little confused on what you mean by the [binary] expression. $\endgroup$
    – mathisfun
    Commented Sep 6, 2021 at 18:37
  • $\begingroup$ @mathisfun Wolfram Alpha factored $1,142,857$ here. I expressed the results in terms of powers of $2$ because the only digits allowed were $4,\space 2,\space 0.\space$ By binary expression, I mean the collection of terms that are powers of $2$. $\endgroup$
    – poetasis
    Commented Sep 6, 2021 at 19:08
  • $\begingroup$ Oh I see! Thanks very much. I will ask you any questions if I am still confused :) $\endgroup$
    – mathisfun
    Commented Sep 6, 2021 at 19:39
  • $\begingroup$ @mathisfun I think you will love Wolfram Alpha as is can do almost anything mathematical (like factoring or expanding or solving polynomial expressions) and many non-mathematical things you can imagine. $\endgroup$
    – poetasis
    Commented Sep 6, 2021 at 20:13
  • $\begingroup$ Oh ok! I'll check it out. Thanks for the advice! $\endgroup$
    – mathisfun
    Commented Sep 7, 2021 at 1:20

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