What is last digit #762^1816# ?

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Answer 1 · 2017-11-09T17:22:15

#6#

Note that powers of #2# have last digit following the repeating pattern:

#2, 4, 8, 6, 2, 4, 8, 6,...#

Also #1816# is divisible by #4# since #100# is divisible by #4# and #16# is divisible by #4#.

So #762^1816# has last digit #6#

Answer 2 · 2017-11-09T17:24:16

#6#

for all numbers whose last digit is #2#, the last digits of their powers have a pattern that repeats for every #4#th integer power:

#2, 4, 8, 6#

examples:

#2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16#

#12^1 = 12, 12^2 = 144#, etc.

#762# also ends in #2#, so it will follow this pattern.

#1816/4 = 454#, so #1816# is a multiple of #4#.

this means that the last digit of #762^1816# will be the fourth term in the sequence.

the last digit of #762^1816# is #6#.