Show, with proof, that the minimum value of $c$ such that $c^n + 2014$ has all digits less than 5, where $c, n \in \mathbb N$, for all values of $n$

Question

Question: Determine, with proof, the minimum value of $c$ such that $c^n + 2014$ has all digits less than 5, where $c, n \in \mathbb N$ for all possible values of $n$. Note that $\mathbb N$ does not include $0$. In other words, $c,n$ shall not assume the value of $0$ or anything lesser than $1$ in general.

For the first $10$ positive integers, we know that

In the following lines, any power refers to any positive power $\geqslant 1$

• $1$ raised to any power always results in $1$, and consequently, the unit digit is also $1$
• $2$ and $8$ raised to any power results in $2$, $4$, $8$, $6$ or $8$ at the unit's place
• $3$ and $7$ raised to any power results in $3$, $9$, $7$ or $1$ at the unit's place
• $4$ raised to any power results in $4$ or $6$ at the unit's place
• $5$ raised to any power results in $5$ at the unit's place
• $6$ raised to any power results in $6$ at the unit's place
• $9$ raised to any power results in $1$ or $9$ at the unit's place

Since we are asked to find the minimum value of $c$, we shall start from 1 and establish certain facts on the way as we progress to greater values of $c$.

The unit digit of whichever value of $c$ we choose should be such that

• it either always ends in $0$ and
• contains no other digit at places other than the unit place which are greater than $3$.

From the above list, we see that none of the single-digit numbers fulfill this requirement. Hence, the immediate smallest positive integer that we have is $10$.

Now, we know $10$ raised to any power always results in $0$ at the unit's digit and the only other digit present in any number $k = 10^n$ is $1$ and even if we add $1$ to any of the digits in $2014$, the condition of having digits $\lt 5$ in the resulting number is maintained. Hence, $10$ is the minimum value of $c$.

This is my proof for the problem stated in the title. Is it correct / self-explanatory / un-assuming ?

Answer 1

The minimal value for the present problem, as it stays after some edits, is indeed $c=10$. To see this, one has to mention counterexamples of $n$ for the cases of $c$ among $1,2,3,4,5,6,7,8,9$, and to show that $c=10$ is indeed a match: $$ \begin{aligned} 2014 + 1 ^1 &=2015\ , &&\text{ rejects }c=1\ ,\\ 2014 + 2 ^1 &=2016\ , &&\text{ rejects }c=2\ ,\\ 2014 + 3 ^1 &=2017\ , &&\text{ rejects }c=3\ ,\\ 2014 + 4 ^1 &=2018\ , &&\text{ rejects }c=4\ ,\\ 2014 + 5 ^1 &=2019\ , &&\text{ rejects }c=5\ ,\\ 2014 + 6 ^2 &=2050\ , &&\text{ rejects }c=6\ ,\\ 2014 + 7 ^2 &=2063\ , &&\text{ rejects }c=7\ ,\\ 2014 + 8 ^2 &=2078\ , &&\text{ rejects }c=8\ ,\\ 2014 + 9 ^2 &=2095\ , &&\text{ rejects }c=9\ ,\\[3mm] &\qquad\text{and for }c=10\\[3mm] 2014 + 10^1 &= 2024\ ,\\ 2014 + 10^2 &= 2114\ ,\\ 2014 + 10^3 &= 3014\ ,\\ 2014 + 10^4 &= 12014\ ,\\ 2014 + 10^n &= 2014+10\cdots0=10\cdots2014\ ,\\ \end{aligned} $$ the last line for all $n\ge 4$, so the value $c=10$ is a solution, and it is the minimal solution.

$\square$

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The minimal value of $c$ for the "other problem" (...find minimal natural $c\ge 1$ so that there exists $n\ge 1$ with $2014+c^n$ having digits among $0,1,2,3,4$) is $2$, since $1$ is not a solution, and for instance: $$ \begin{aligned} 2014 + 2^3 &= 2022\ ,\\ 2014 + 2^4 &= 2030\ ,\\ 2014 + 2^7 &= 2142\ . \end{aligned} $$

Answer 2

Yes, your answer is correct, but it can be simplified.

It is easy to verify that $10$ works (assuming that $\mathbb N$ is the positive integers - nothing works if $n$ can be $0$).

Now $1$ to $5$ don't work because taking $n=1$ we have a value between $2015$ and $2019$ - the units digit is too big.

Also, $6$ to $9$ don't work, because taking $n=2$ we have a value between $2050$ and $2095$ - the tens digit is too big.