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#

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" "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#

Answer 2 · 2018-03-13T22:14:14

#2#

powers of #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#.