# Thread: The last digit of 7^7^7

1. ## The last digit of 7^7^7

I've used modular exponentiation on $7^7$ 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)

2. ## Re: The last digit of 7^7^7

Originally Posted by Celcius
I've used modular exponentiation on $7^7$ 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

3. ## Re: The last digit of 7^7^7

Originally Posted by TheChaz
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?

4. ## Re: The last digit of 7^7^7

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.

5. ## Re: The last digit of 7^7^7

Hello, Celcius!

$\text{Find the last digit of }7^{7^7}$

Be careful!

This is not $(7^7)^7 \:=\:(823,543)^7$ . . . which has only 42 digits.

Given a "stack" of exponents and no parentheses, we read "from the top down".

The number is: . $7^{(7^7)} \:=\:7^{823,543}$ . . . which has nearly 700,000 digits.

Consider the last digits of powers-of-7:

. . $\begin{array}{ccc} 7^1 & \to & 7 \\ 7^2 & \to & 9 \\ 7^3 & \to & 3 \\ 7^4 & \to & 1 \\ 7^5 & \to & 7 \\ \vdots && \vdots \end{array}$

We see that the last digits form a 4-step cycle: . $7, 9, 3, 1$

We find that: . $7^7 \:=\:823,543 \:=\:4(205,885) + 3$

Therefore, the last digit of $7^{7^7}$ is the same as $7^3$: .3

6. ## Re: The last digit of 7^7^7

Originally Posted by Soroban
Hello, Celcius!

Be careful!

This is not $(7^7)^7 \:=\823,543)^7 " alt="(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: . $7^{(7^7)} \:=\:7^{823,543}$ . . . which has nearly 700,000 digits.

Consider the last digits of powers-of-7:

. . $\begin{array}{ccc} 7^1 & \to & 7 \\ 7^2 & \to & 9 \\ 7^3 & \to & 3 \\ 7^4 & \to & 1 \\ 7^5 & \to & 7 \\ \vdots && \vdots \end{array}$

We see that the last digits form a 4-step cycle: . $7, 9, 3, 1$

We find that: . $7^7 \:=\:823,543 \:=\:4(205,885) + 3$

Therefore, the last digit of $7^{7^7}$ is the same as $7^3$: .3

Thanks for this answer, the only thing is how do I get the deduction that $7^{7^7}$ is the same as $7^3$?

7. ## Re: The last digit of 7^7^7

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).

8. ## Re: The last digit of 7^7^7

This problem is not hard.
We see $7 \equiv -1 \pmod{4} \Rightarrow 7^7 \equiv -1 \pmod{4}$.
Then let $7^7=4k+3$.
So $7^{7^7}=7^{4k+3}=(7^4)^k.7^3$.
From here it is easy to find next.

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# find last two digits of 7^7^7

Click on a term to search for related topics.