How do you find the remainder of 3^983 divided by 5?

Question

Julia Adams · Accepted Answer

A remainder of #2#

You are obviously not going to work out the actual value of #3^983#!!

Let's look at the pattern of the powers of #3#

#3^1 = 3# #3^2=9# #3^3 = 27# #3^4 = 81# #3^5 =243# #3^6=729#

You can observe a pattern that repeats over four numbers if you look at the final digit.

#3,9,7,1" "3,9,7,1" "3,9,7,1#

So to find out what the last digit of #3^983# is, divide by #4#

#983 div 4 = 245 3/4#

This means it is the third number in the pattern of #4#, so the last digit will be a #7#

Therefore when you divide that number by #5# there will be a remainder of #2#

Ethan Adkins · Answer

#2#

powers of #3:#

#3,9,27,81,243,729...#

There's a pattern in their final numbers:

#3,9,7,1(,3,9...)#

Every fourth power, it repeats.

e.g. powers of #3# that are multiples of #4# (#3^4, 3^8,# etc.) all have #1# as their last digit.

#983 = 3 + 980#

#983# is #3# more than a multiple of #4#.

this means that the last digit of #3^983# corresponds to the third term in the sequence, which is #7#.

all multiples of #5# end in either #5# or #0#.

#7# is closer to #5# (than to #10#), and is #2# more than #5#.

any integer ending in a #7# has a remainder #2# when divided by #5#.

since #3^983# ends in #7#, #3^983# has a remainder of #2# when divided by #5#.

Hazel Anglin · Answer

undefined To find the remainder of $3^{983}$ divided by 5, we can use the concept of modular arithmetic and Euler's theorem. Euler's theorem states that if $a$ and $n$ are coprime integers (i.e., they have no common factors other than 1), then $a^{\phi(n)} \equiv 1 \pmod{n}$, where $\phi(n)$ is Euler's totient function, which gives the number of positive integers less than $n$ that are coprime to $n$.

In this case, since 5 is prime, $\phi(5) = 5 - 1 = 4$. So, according to Euler's theorem:
$$3^{4} \equiv 1 \pmod{5}$$

Now, we need to find the remainder of $3^{983}$ divided by 4:
$$983 \equiv 3 \pmod{4}$$

So, $3^{983} \equiv 3^3 \pmod{5}$:
$$3^{983} \equiv 27 \pmod{5}$$
$$3^{983} \equiv 2 \pmod{5}$$

Therefore, the remainder of $3^{983}$ divided by 5 is 2.