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

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Answer 1 · 2018-03-13T22:08:00

A remainder of $2$ $2$

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

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

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

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

$3, 9, 7, 13, 9, 7, 13, 9, 7, 1$ $3,9,7,1" "3,9,7,1" "3,9,7,1$

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

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

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

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

Answer 2 · 2018-03-13T22:14:14

$2$ $2$

powers of $3 :$ $3:$

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

their last digits have a pattern:

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

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 $5$ .