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

Mar 13, 2018

#### Answer:

A remainder of $2$

#### Explanation:

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

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

${3}^{1} = 3$
${3}^{2} = 9$
${3}^{3} = 27$
${3}^{4} = 81$
${3}^{5} = 243$
${3}^{6} = 729$

If you look at the last digit you will see a pattern that repeats over four numbers..

$3 , 9 , 7 , 1 \text{ "3,9,7,1" } 3 , 9 , 7 , 1$

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

$983 \div 4 = 245 \frac{3}{4}$

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 $2$

Mar 13, 2018

#### Answer:

$2$

#### Explanation:

powers of $3 :$

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

their last digits have a pattern:

$3 , 9 , 7 , 1 \left(, 3 , 9. . .\right)$

it repeats for every fourth power.

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 $5$.