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I was going through this answer: https://math.stackexchange.com/a/1852190/989402

I found the answer was correct, but not satisfied with the solution. I mean if we have $2^{10n+1}$ and n is even will always yield a last digit of $2$, and if n is odd will yield a last digit of $8$ always. And since the question was $2^{9^{100}}$ will yield last digit of $2$ only when $9^{100}$ yields even digit at second place and first place will be $1$ (or $5$ which isn't possible), which it is.

So in short my question is that how will we prove that $9^n$ yields always a even digit at second place?

EDIT: Assuming number abcdefg, f is the second digit for this question.

Wild Martian
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    Do you mean second-to-last digit? Because e.g. the second base-$10$ digit of $9^2$ is $1$. – J.G. Dec 31 '21 at 08:59
  • No I actually mean the $10^{th}$ place. – Wild Martian Dec 31 '21 at 09:04
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    You mean the tens place, so yes, second-to-last. – J.G. Dec 31 '21 at 09:05
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    Hint: $9^n$ has last digit either $1$ or $9$. Suppose $9^n = \text{something even} \times 10 + c$ (where $c \in {1,9}$), then show that $9^{n+1}$ can be written in this form as well – Ray Dec 31 '21 at 09:08
  • Another approach using Ray's hint is to consider the residue mod 4 of a number $u1$ or $u9$ , where $u$ is an od digit and also to notice that $9^n$ is a perfect square for all positive integers $n$. – Peter Dec 31 '21 at 09:33
  • Special case of the first linked dupe. More generally see here in the 2nd linked dupe for a closed form for the tens and units digits. – Bill Dubuque Dec 31 '21 at 11:44

4 Answers4

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The class of numbers of the form $10x+y$ where $x$ is even and $y$ is $1$ or $9$ is closed under multiplication and $9$ belongs to it, hence $9^n$ always belongs to that class.

To see this note $(10x+y)(10u+v) = 10(10xu + xv+uy) + yv$ where $10xu+xv+uy$ is a sum of $3$ even numbers. The case in which $yv$ is $1$ or $9$ is immediate, and in the case $yv=81$ the expression can be written as $10(10xu+xv+uy+8)+1$.

Asinomás
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Take the powers of $9$ modulo $20$. The sequence goes $\{1,9, 1 \dots \}$.

Explicitly, we can write: $$9^{2k} = 20 p + 1\\\ 9^{2k+1} = 20 p + 9$$

Numbers of that form will have the second-to-last digit equal to $2p$ $\text{mod}$ $10$ (the last digit of $2p$), which is even.

user3257842
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a brute force approach is to list out all the powers of 9 modulo 100.
9,81,29,61,49,41,69,21,89,1
then verify second digit is even for all the 10 residues.
taking modulo 20 is faster though.

cineel
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One can show that the odd decimal numbers ending in 1, 3, 7, 9, and an even number of digits, has an even number of factors with an odd tens digit,

The proof for this is in Gauss's quadratic reciprocality test, for a base of -5. Relative to 5, squares have a remainder of 1, 4 mod 5. Relative to -1, squares must end in 1, mod 4. Takem together, the quadratic residue is even when there is an odd number of odd values, so

  1  3  7  9  11  13  17  19
  e  o  o  e   e   o   o   e    by rule for 5, 1 and 4 even.
  e  o  o  e   o   e   e   o    by rule of -1, 1 mod 4 even.
  e  e  e  e   o   o   o   o    by rule of -5,  tens digit odd.

This means that the product of any number of primes not dividing 10, which have an even tens-digit will have an even tens digit. This includes for example, numbers of the form $3^a 7^b$ for all a, b.

Since 9 is a member of this set, all powers of 9 have an even tens digit.

wendy.krieger
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