# What is the lowest digit in n or in 7n?

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 $m$?

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 $m\le 6.$ $\blacksquare$

• +1 Because i like how this answer addresses the issue of the first digit directly, and concisely. Jul 12, 2017 at 12:46
• 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!) Jul 12, 2017 at 12:48
• If you don't mind, how did you come up with this idea?
– Ovi
Jul 13, 2017 at 4:14
• @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. Jul 13, 2017 at 4:34
• 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.
– Ovi
Jul 13, 2017 at 4:47

$m$ 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.

6.

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.

• @Ankoganit Fair point. Jul 12, 2017 at 12:49

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.