I've used modular exponentiation on $\displaystyle 7^7$ to find the last digit is 3.
How can I use that fact to find the last digit of 7^7^7? (7 to the power of 7, to the power of 7)
I think you need more than (just) the last digit of 7^7.
Notice that 7 = 7 (mod 10)
7^2 = 49 = -1 (mod 10)
7^3 = 7*-1 (mod 10) = -7 (mod 10)
7^4 = 1 (mod 10)
7^5 = 7 (mod 10)... now we have repeated.
So you need to know what 7^7 is (mod 4), since the pattern is length 4. This will require the last TWO digits of 7^7, which are 43 = 3 mod 4. So the last digit of 7^(7^7) should be -7 = 3.
I've tried to color-code a little so you can see the train of thought
Not sure, I completely understand what you are saying, also I am colour blind, so I don't think I get that part either.
Ok, so my first part, was actually using modular exponentiation to find the answer to 7^7 mod 4 = 3
I know I said last digit, but I mixed that part up with the second part.
But, I found that the answer is 7^7 mod 4 = 3. Now how does that fact tie into the second part?
7^1: 7 (mod 10) = 7
7^2: 49 (mod 10) = 9
But what is the -1? And the next line 7^3, how did you arrive at that?
7^1 = 7
7^2 = 49 = 9 = -1 (mod 10). Using negative numbers isn't necessary.
7^3 = 49*7 = 9*7 = -1*7 (mod 10) ... here is just multiplies the entire previous line by 7. This simlpifies to
7^3 = 343 = 63 = -7 (mod 10). Notice that this also equals 3 (mod 10).
Then 7^4 = 7 times any/all of those = 1 (mod 10
7^5 = 7 (mod 10)
etc.
Since the LAST digit of powers of 7 follows this pattern, we need to know what the exponent of 7 is, mod 4. The exponent is ....43 = 3 mod 4, so
"The last digit of 7^(7^7) is the same as the last digit of 7^43 and of 7^3. "
3.
Hello, Celcius!
$\displaystyle \text{Find the last digit of }7^{7^7}$
Be careful!
This is not $\displaystyle (7^7)^7 \:=\:(823,543)^7 $ . . . which has only 42 digits.
Given a "stack" of exponents and no parentheses, we read "from the top down".
The number is: .$\displaystyle 7^{(7^7)} \:=\:7^{823,543}$ . . . which has nearly 700,000 digits.
Consider the last digits of powers-of-7:
. . $\displaystyle \begin{array}{ccc} 7^1 & \to & 7 \\ 7^2 & \to & 9 \\ 7^3 & \to & 3 \\ 7^4 & \to & 1 \\ 7^5 & \to & 7 \\ \vdots && \vdots \end{array}$
We see that the last digits form a 4-step cycle: .$\displaystyle 7, 9, 3, 1$
We find that: .$\displaystyle 7^7 \:=\:823,543 \:=\:4(205,885) + 3$
Therefore, the last digit of $\displaystyle 7^{7^7}$ is the same as $\displaystyle 7^3$: .3
Really? We've done this twice!
The last digit ("last digit" means mod 10) of powers of 7:
7
9
3
1
7
9
3
1
.
.
.
7^3 has last digit 3
7^7 has last digit 3
7^11
7^15
7^19
.
.
.
7^10000000000000000000000000000000003 will also have last digit equal to three, since the exponent is congruent to 3 (mod4).