Tower of 7s

Looking at small powers of $7$ , notice that we only care about the last digit, so we can easily compute the last digit of $7^n$ :

$\begin{aligned} 7^1&\equiv7\\ 7^2&\equiv7\cdot7\equiv9\\ 7^3&\equiv7\cdot9\equiv3\\ 7^4&\equiv7\cdot3\equiv1\\ 7^5&\equiv7\cdot1\equiv7\\ 7^6&\equiv7\cdot7\equiv9\\ 7^7&\equiv7\cdot9\equiv3\\ 7^8&\equiv7\cdot3\equiv1 \end{aligned}$

This suggests the pattern that $7^{4k}\equiv1$ , $7^{4k+1}\equiv7$ , $7^{4k+2}\equiv9$ , and $7^{4k+3}\equiv3$ . Using modular arithmetic (thank you Calvin Lin), we get $7^{7^{7^{7^{7^7}}}}=4k+3$ , so the answer is $\boxed{3}$ .

Let 7(n) be a tower of n 7's. Examine the powers of 7 under modulo 4:

7 = 7;

7^2 = 49 = 1.

So 7^a = 7 and 7^b = 7 for odd a and even b. Since 7(5) is odd, so 7(6) = 7^(7(5)) = 7 (mod 4). Examine the power of 7 under modulo 10:

7 = 7;

7^2 = 9;

7^3 = 3;

7^4 = 1.

Therefore 7(7) = 7^(7(6)) = 7^7 = 7^3 = 3 (mod 10). The final digit is 3.

The answer seems to be wrong. It should be $7$ . My argument is as follows. Let's check the first $n^7$ , where $1 \le n \le 30$ is an integers . They are as follows:

$1^7= 1$

$2 ^7=128$

$3^7= 2187$

$4 ^7=16384$

$5^7= 78125$

$6^7= 279936$

$7^7= 823543$

$8^7= 2097152$

$9^7= 4782969$

$10^7= 10000000$

$11^7= 19487171$

$12^7= 35831808$

$13^7= 62748517$

$14^7= 105413504$

$15^7= 170859375$

$16^7= 268435456$

$17^7= 410338673$

$18^7= 612220032$

$19^7= 893871739$

$20^7= 1280000000$

$21^7= 1801088541$

$22^7= 2494357888$

$23^7= 3404825447$

$24^7= 4586471424$

$25^7= 6103515625$

$26^7= 8031810176$

$27^7= 10460353203$

$28^7= 13492928512$

$29^7= 17249876309$

$30^7= 21870000000$

It can be seen that the last digit if $n^7$ depends only the last digit of $n$ . They are in a sequence of $1, 8, 7, 4, 5, 6, 3, 2, 9, 0$ . Therefore,

$7^7 \equiv 3 \mod {10}$

$(7^7)^7 \equiv 7 \mod {10}$

$((7^7)^7)^7 \equiv 3 \mod {10}$

$(((7^7)^7)^7)^7 \equiv 7 \mod {10}$

$((((7^7)^7)^7)^7)^7 \equiv 3 \mod {10}$

$(((((7^7)^7)^7)^7)^7)^7 \equiv 7 \mod {10}$

The answer should be $\boxed{7}$ and not $3$ . I guessed the 'wrong' answer.

very Simple Solution.. Came from me... XD

The answer is 3:

7^7^7^7^7^7^7

Evaluate: 7^7 = 823543

Since the cycle is always: 7^1 = 7 7^2 = 9 (last digit of 49) 7^3 = 3 7^4 = 1

Divide 823543 to 4, youll get: 205885 remainder 3

So, it's last digit is on 3..

7^7^7^7^7^823543

Since 7^823543 ends in 3

7^7^7^7^(...3) , where in, (...3) is equivalent to the value of 7^823543

Again, 3 is the last digit of 7^(...3)

So, until to the last 7, the last digit is 3..

There fore, the last digit of 7^7^7^7^7^7^7 is 3...

Thanks to Calvin Lin! I can correct my explanation.

The last digit of N = (7 ↑↑ 7) is: LD(7 ↑↑ 7) = 3.

Proof:

7^(7^(7^(7^(7^(7^7))))) = 7 ↑↑ 7; (tetration of "7" "7's")

7^0 => LD(7^0) = 1 = LD(7^(4n); => 7^(4n) ≡ 1 mod 10;

7^1 => LD(7^1) = 7 = LD(7^(4n + 1)); => 7^(4n + 1) ≡ 7 mod 10;

7^2 => LD(7^2) = 9; LD(7^(4n + 2)); => 7^(4n + 2) ≡ 9 mod 10;

7^3 => LD(7^3) = 3 = LD(7^(4n + 3)); => 7^(4n + 3) ≡ 3 mod 10;

there is a periodicity with period c = 4, n ∈ ℕ ∪ {0};

Since, because:

LD(7 ↑↑7) = 7 ↑↑ 7 mod 10, with period c = 4;

7 ≡ 3 mod 4; but also:

7 ≡ -1 mod 4;

so, because:

4k - 1 = 4(k - 1) + 3, is odd, we have:

7^(4(k - 1)+ 3) ≡ (- 1)^(4(k - 1)+ 3) mod 4,

7^(4(k - 1) + 3) ≡ - 1 mod 4,

so:

7^(4(k - 1) + 3) = 4N + 3;

so, LD(7^(4(k - 1) + 3) = 3, and so:

(7 ↑↑ 2) = 7^7 = 7^(4N2 + 3) = (4N3 + 3); N2 = 1

(7 ↑↑ 3) = 7^(4N3 + 3) = (4N3 + 3); N3 = 205885;

(7 ↑↑ 4) = 7^(4N4 + 3) = (4N5 + 3) N4 = ((7 ↑↑ 3) - 3)/4

(7 ↑↑ 5) = 7^(4N5 + 3) = ...

(7 ↑↑ 6) = 7^(4N6 + 3) = ...

(7 ↑↑ 7) = 7^(4N7 + 3) = ...

(7 ↑↑ n) = 7^(4Nn + 3) = ...

k, n ∈ ℕ - {1}; N, Nk ∈ ℕ;

so, ANSWER:

LD(7 ↑↑ 7) = 3,

and: LD(7 ↑↑ n) = 3

7 7 7 7 7 7 7 =823543