# Finding the final digit

What is the final digit in $\left(\cdots \left(\left(\left(7^7\right)^7\right)^7\right)^7\cdots\right)^7$, where the $7^{\text{th}}$ power is taken $100$ times?

So I'm trying to see how to do this problem without a calculator but i dont think a calculator would help because the numbers would be too big.

There is probably a trick but i can't see it.

In the calculator, I did $7^7=823543$

and i did $823543^7=2.5692358\times 10^{41}$

Since exponential is being involved, i would guess the final digit would be $0$ but i'm not sure.

Any explanations?

• $7^1=7$, $7^2 =49$, $7^3 = 343$, $7^4 = 2401$, $7^5 = 16,807$, $7^6 = 117,649$, $7^7 = 823,543$, $7^8 = 5,764,801$. Notice the pattern on the last two digits? – Ben Oct 4 '15 at 20:28
• The final digit can only be $0$ if the integer is divisible by $2$ and $5$. In this case it's only divisible by the prime $7$. – user236182 Oct 4 '15 at 20:37

Think modulo $10$:

The powers of $7$ modulo $10$ are, in order, $1,7,9,3$, and then it repeats. Specifically, $7^7\equiv_{10} 3$.

Now, the powers of $3$ modulo $10$ are $1,3,9,7$, and then it repeats. Specifically, $3^7\equiv_{10}7$.

We see that for each time we take the seventh power, the last digits alternates between seven and three. We start with $7$ and take the seventh power $100$ times, that is, an even number of times. We end up back at seven, which is the last digit.

The expression is equal to $7^{7^{100}}$. Since

$7^{4}\equiv \left(7^2\right)^2\equiv (-1)^2\equiv 1\pmod{10}$, we get

$7^{7^{100}}\equiv 7^{7^{100}\bmod {4}}\equiv 7^{(-1)^{100}}\equiv 7^{1}\equiv 7\pmod{10}$

Last digit for $7^7 = 823543$ is $3$

Last digit for $(7^7)^7$ is $7$

Last digit for $((7^7)^7)^7$ is $3$

....

Last digit for $7^{th}$ power taken odd number of times is $3$

Last digit for $7^{th}$ power taken even number of times is $7$

Last digit for $7^{th}$ power taken $100$ times should be $7$

All the ten digits $1,2,....9,0$, have its powers admiting a period $4$ modulo $10$ (also periods $2$ and $1$ so period $4$ too) For $7$ we have the rests $7,9,3,1$ so $7^7$ has rest $3$; for $3$ we have the rests $3,9,7,1$ so $3^7$ have rest $7$ thus $7^{7^7}$ have rest $7$. Since $\frac{100}{50}=2$ the searched rest is $\boxed7$.

Im just giving a different way .According to Euler's maths prime numbers such as $3,7$ raised to some powers show same patterns when powers change by numbers such as $2,4,6,8..100$ then the last digit again becomes same Eg (last digit)$7^3=7^7$ last digit is $3$ in $343$ and $823543$ so our problem gets reduced to $(7^3)^3$..100 times now when we do $(7^3)^3$ we get $40353607$ last digit is $7$ so the last digit of your original expansion would also be $7$. of course you can always do modulo but for numbers such as $7,3$ you can think like this also. hope its clear.