Find the last digit of $43^{43^{43}}$

Question

Find the last digit of $43^{43^{43}}$. My attempt:

$$(40+3)^3 \equiv 3\cdot 40\cdot 9+27 \equiv 1080+27\equiv 07\pmod{100}$$

Am I right or is there any other way to find the answer?

Answer 1

Hint. To find the last digit of $43^{43^{43}}$ it suffices to work modulo 10. Note that $43^{4n}\equiv 3^{4n}\equiv 1 \pmod{10}$. Hence we need to find the exponent $43^{43}$ modulo 4: $43^{43}\equiv (-1)^{43}=-1\equiv 3\pmod{4}$.

Can you take it from here?

P.S. Finally, you should find: $$43^{43^{43}}\equiv 3^{4n+3}=3^{4n}\cdot 3^3\equiv 1\cdot 27\equiv 7\pmod{10}.$$

Answer 2

$$\color{red} {a^{4k+r}\equiv a^r (mod 10)\\r=1,2,3,4}$$ so find $43^{43}=4k+r $in mod $4$ then $43^{43^{43}}$ in mod $10$ $$43^{43} \equiv (40+3)^{43}(mod 4)\\\equiv (+3)^{43}(mod 4)\\\equiv (-1)^{43}(mod 4)\\\equiv (+3)(mod 4)$$ now $$43^{43^{43}} \equiv (43)^{4k+3}(mod 10)\\ \equiv (43)^{3}(mod 10)\\ \equiv (40+3)^{3}(mod 10)\\ \equiv (3)^{3}(mod 10)\\ \equiv 27(mod 10)\\\equiv 7(mod 10)\\$$

Answer 3

Finding last digit $\iff$ $\pmod{10}$

Now as $43\equiv3,43^n\equiv3^n\pmod{10}$

Now $43^{43}$ can be written as $(4a-1)^{2b+1}\equiv-1\pmod4$ which can be expressed as $4c+3$ where $a,b,c$ are positive integers.

Now $3^{4c+3}=3^3\cdot(1+80)^c\equiv7\pmod{10}$

Answer 4

There will come some number $n $ where the last digit of $43^n $ is $1$.( $43^{n}\equiv 1 \mod 10$) Then any $43 ^{kn+m}=(43^n)^k43^m\equiv 1^k43^m \mod 10$ which will probably be easier to solve.

So our first step is to find that $n $.

Since $43=40+3$ then $43^n=40^n+a_140^{n-1}*3+......+a_{n-1}40*3^{n-1}+3^n $ and all but the last term is a multiple of $10$. So $43^n $ will have the same last digit as $3^n $.

Now there is something called Euler's theorem that will tell me exactly what the $n$ so that $3^n\equiv 1 \mod 10$. But I'll assume you don't know it.

Instead we'll try it till we get it.

$3^2=9\equiv -1\mod 10$ so $3^4\equiv (-1)^2\equiv 1 \mod 10$. So $n=4$ was the number we wanted: $43^{4k+m}\equiv 3^{4k+m}\equiv (3^4)^k*3^m\equiv 1^k*3^m\equiv 3^m\mod 10$.

So the last digit of $43^{43} $ is the same last digit of $3^3=27$ which is $7$.

But that does't help us solve $43^{43^{43}} $.

We need to now how $43^{43}=4k+m $ is so the last digit of $43^{43^{43}}=43^ {4k+m}\equiv 3^m\mod 10$.

So that's the next thing we need to figure out. What is $m$ if $43^{43}=4k+m $. (Note:we don't care what $k $ is.)

$43= 40+3$ so $(40+3)^s $ will be $40^s+......+3^s $. Just like before, only the last term is not a multiple of $4$ so $43^s\equiv 3^s \mod 4$.

Like above $3^2=9\equiv 1\mod 4$ so $3^{2k+i}\equiv (3^2)^k 3^i\mod 4$.

So $43^{43}\equiv 3^{43}\equiv (3^2)^{21}*3\equiv 3\mod 4$.

So $43^{43}=4k+3$ for some $k $.

So $43^{43^{43}}\equiv 3^{4k+3}\equiv 3^3\equiv 27 \equiv 7\mod 10$.

So the last digit is $7$.