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For any positive integer $n$, consider the digits which occur either in $n$ or in $7n$. Let $m$ be the smallest digit among those digits. What is the largest possible value of $m$?

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The largest possible value of $m$ is

$6$.

Proof:

The value $6$ can be attained, for say, $n=98$, so that $7n=686$. Now suppose $n$ has $k$ digits and each digit is at least $7$. Then $7\times 10^{k-1}\le n<10^{k}\implies 4.9\times 10^k\le 7n<7\times 10^k$, so $7n$ starts with a digit strictly between $3$ and $7$, which means $m\le 6.$ $\blacksquare$

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    $\begingroup$ +1 Because i like how this answer addresses the issue of the first digit directly, and concisely. $\endgroup$ – Piotr Pytlik Jul 12 '17 at 12:46
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    $\begingroup$ Agreed, this is simpler than my proof. (I assumed starting from the right would be easier, because everything higher just carries over, but this is very neat!) $\endgroup$ – Rand al'Thor Jul 12 '17 at 12:48
  • $\begingroup$ If you don't mind, how did you come up with this idea? $\endgroup$ – Ovi Jul 13 '17 at 4:14
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    $\begingroup$ @Ovi the answer can be guessed via checking some potential candidates. For the proof, I first started reasoning about the last digits, but that quickly got casework-y, so I tried the first digits instead. $\endgroup$ – Ankoganit Jul 13 '17 at 4:34
  • $\begingroup$ Thanks for the reply. This serves to reinforce a lesson which I've learned recently: try going against intuition. If the problem asks for the shortest, look at the longest, if the problem asks for the first, look at the last, etc. $\endgroup$ – Ovi Jul 13 '17 at 4:47
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$m$ can definitely be at least

$6$: for example, let $n=97$ so that $7n=679$, giving $m=6$.


It cannot be larger than this, because:

if $m\geq7$, then $n$ and $7n$ each consists only of the digits $7,8,9$.

The final digit of $n$ must therefore be

$7$, because otherwise the final digit of $7n$ will be $6$ or $3$, contradiction.

Similarly, the second-to-last digit of $n$ must be

$9$, because otherwise the last two digits of $7n$ will be $39$ or $09$ (these figures got by calculating $7\times77$ and $7\times87$).

Then the third-to last digit of $n$ must be

$9$, because otherwise the last three digits of $7n$ will be $579$ or $279$ (these figures got by calculating $7\times797$ and $7\times897$).

And so on, by induction. We get that $n$ must be of the form

$n=99...997$, giving $7n=699...9979$, contradiction.


So the answer is

6.

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Proof:

For any number n, when the leftmost digit is multiplied by 7 in the case of 7n, the new leftmost digit cannot be higher than 6, as the original leftmost digit cannot be greater than 9 (9 times 7 is 63). The case of 97 times 7 equalling 679 shows that 6 is achievable, and the above disproves the possibility of 7 or greater being possible.

Further proof as requested:

I stated that the leftmost digit cannot be greater than 6, even when carrying forward digits. The largest digit on the left (without carrying anything) is a 6, created by multiplying 9 by 7 (63). For this 6 to become a 7, the 3 must have at least 7 added to it. However this is not possible, as the largest number that can be created in the tens position by multiplying 7 by a single digit number is a 6 (9 times 7 is 63, a 6 is in the tens position). Hence, only 6 would be carried over in the most extreme example, leaving 6 in the leftmost position. This can be seen when multiplying 7 by an infinite chain of 9s (999999999...) - this would create the number 69999999999999....3.

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I think it's

6.

Proof (sort of)

$97 \times 7 = 679$ results in $m = 6$
This means that the value we are looking for is 6 or higher.
Let's try to find a number n that will lead us to m>6.
The number n must end in 7 otherwise we will get the last digit of $n$ or $7\times n$ be 6 or lower.
1 to 6 will mean m<=6.
8 results in last digit of 7n be 6.
9 results in last digit of 7n be 3.
For first digit of n be 8 or lower we get m be 6 or lower.
so first digit of n must be 9.
n looks like $9....7$ where the dots can be 7, 8 or 9.

Case 1.

All digits are 9.
this means $n = 10^k - 3$.
$7 \times n = 7 \times 10 ^k - 21$ which starts with a 6, has k-2 nines after and 79 at the end. So m = 6.

Case 2

if n is 9....87 then n = $n = a \times 10^{k} - 13$.
$7 \times n = 7\times a \times 10^{k} - 91$ which ends with 09 so we got a 0. not cool.
in the same matter we show that 9....77 ends with 39. Again, not cool.

Still thinking of the case when 8 or 7 are in the middle of n.

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