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 $n$

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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 ?

Best Answer

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.

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