Finding Tens Digit of Very Large Number I think I can figure out the leading digit or the units digit, but how would one find the tens digit of, say, $$7^{100}$$ without having a calculator or web app that actually displays the full integer.
Is there a way to find the tens digit given a scientific calculator that can only display the answer in scientific notation?
Also, I have no idea what to tag this question.
 A: The tens digit of $n$ can be read off of $n\pmod {100}$.  In this case, $7^4\equiv 1 \pmod {100}$ so $7^{100}\equiv 1 \pmod {100}$, so the answer is $0$.
A: $7^{100}$ turns out to be a bit boring, so let's work up a method for finding the last two digits of $13^{37}$ with your calculator. Everyone else has given results using modular arithmetic, so I'll just make a basic explanation/exploration of what is possible.
$13^2 = 169$. Whatever we multiply this by, the $100$ portion is not going to affect the last two digits, so we can throw that away in any further calculation and just focus on the $69$. This is true for every calculation, so we could just repetitively multiple by $13$, subtract off the hundreds, do it again:
$$13\cdot 13 = 169 \to 69\cdot 13 = 897 \to 97\cdot 13 = 1261 \to 61\cdot 13 = 793\\ \to 93\cdot 13 = 1209 \to 9\cdot 13 = 117 \to 17\cdot 13 = 221 \to 21\cdot 13 = 273 \to 73$$
So here we've got as far as finding that the last two digits of $13^9$ are $73$.
We don't have to go one step at a time, though. The last two digits of $13^6$ were $09$, which would mean that the last two digits of $13^{12}$ will be $09^2=81$. Let's just restart the process above to check that:
$$73\cdot 13 = 949 \to 49\cdot 13 = 637 \to 37\cdot 13 = 481 \to 81$$
Yes. Similarly we can get the last two digits of $13^{18}$ from $09\cdot 81 = 729 \to 29$, and the last two digits of $13^{36}$ from $29^2 = 841\to 41$. Then the last two digits of $13^{37}$ can be found with $41\cdot 13 = 533\to 33$.
With $7^{100}$, which I said was boring, the corresponding first few steps gives us 
$$7.\cdot 7 = 49 \to 49.\cdot 7 = 343 \to 43\cdot 7 = 301 \to 01
$$
and since the last two digits of $7^4$ are $01$, we can immediately say that $01^{25}=01$ and thus the tens digit of $7^{100}$ is zero.
A: Suppose we have a number in the decimal number system with the following representation:
$a_na_{n-1}\ldots a_1a_0$, where all the $a_i \in \{0,1, \ldots, 9\}$. 
This means that the number is 
$$a_n\cdot 10^n + a_{n-1}\cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0\cdot 10^0$$
so if you want to find the digit $a_1$, then you have to reduce the above number modulo 100 (since this will leave you with $a_1 \cdot 10 + a_0$).
Looking at your number, we want to compute $7^{100} \mod 100$. Because $7^4 = 2401$, we have that $7^4 \equiv 1 \mod 100$ and hence we find that
$$7^{100} \equiv 7^{4 \cdot 25} \equiv (7^4)^{25} \equiv 1^{25} \equiv 1 \mod 100.$$
So that $a_1\cdot 10 + a_0 = 1$ and hence $a_1 = 0, a_0 = 1$. So the second last digit is $0$.
Now let me ask you a question: if I were to ask you how to find the '100' digit, how would you do this? 
A: To find unit digit we use mod 10.
To find tens digit mod 100 is used.
$7^4 \equiv 1 \pmod {100}$
$7^{100} \pmod{100} \equiv (7^4)^{25} \pmod {100} \equiv 1^{25} \pmod {100} \equiv 0$
So answer is zero.
