# Thread: divisibility

1. ## divisibility

Find the last digit of the number 2^555666777

2. $\displaystyle 2^1 \equiv 2 \pmod{10}$
$\displaystyle 2^2 \equiv 4 \pmod{10}$
$\displaystyle 2^3 \equiv 8 \pmod{10}$
$\displaystyle 2^4 \equiv 6 \pmod{10}$
$\displaystyle 2^5 \equiv 2 \pmod{10}$
$\displaystyle 2^6 \equiv 4 \pmod{10}$
$\displaystyle 2^7 \equiv 8 \pmod{10}$
$\displaystyle 2^8 \equiv 6 \pmod{10}$

Do you get the idea ?

3. no i dont understand the mod 10 part...we have not learned that in class yet.

so would my number end in 2 since the exponent is divisible by 9?

4. Okay.

You are looking for the last digit of $\displaystyle 2^{555666777}$. You don't need all the other digits. So, you can consider your number modulo $\displaystyle 10$ (which means you only consider the remainder of the number when divided by $\displaystyle 10$). For instance, $\displaystyle 13 \equiv 3 \pmod {10}$ because $\displaystyle 13$ divided by $\displaystyle 10$ leaves a remainder of $\displaystyle 3$. And this gives us the last digit.

Now what I've shown you in my preceding post basically tells us that given $\displaystyle 2^a$ :

- if $\displaystyle a \equiv 0 \pmod{4}$ then the last digit of $\displaystyle 2^a$ is 6.
- if $\displaystyle a \equiv 1 \pmod{4}$ then the last digit of $\displaystyle 2^a$ is 2.
- if $\displaystyle a \equiv 2 \pmod{4}$ then the last digit of $\displaystyle 2^a$ is 4.
- if $\displaystyle a \equiv 3 \pmod{4}$ then the last digit of $\displaystyle 2^a$ is 8.

And in your case, $\displaystyle 555666777 \equiv 1 \pmod{4}$, and therefore the last digit of $\displaystyle 2^{555666777}$ is $\displaystyle 2$.

Does it make sense ?

5. so i need to take 555666777/10 and since my remainder is 1, the last digit of my number ends in a 2. yes i do understand now thank you so much.

another question for clarification on my part. if i were taking 11^597637 and dividing it by 4, would the remainder of this always be 3, no matter what the exponent is?

6. so i need to take 555666777/10 and since my remainder is 1
No ! 555666777/4 gives a remainder of 1.

if i were taking 11^597637 and dividing it by 4, would the remainder of this always be 3, no matter what the exponent is?
The method I exposed here only works for some particular cases (here 2) because of special properties. But in the example you just gave above, using Euler's Theorem, given that $\displaystyle \varphi{(4)} = 1$, we can conclude that 11 raised to any power will give a remainder of 3 when divided by 4. But again, this is a little more advanced !

I encourage you to learn modulus (congruences) and all associated theorems asap, it really helps a lot in number theory.

7. okay thank you