What is the last digit of 3^52?Explain several different ways this could've been done.

What is the last digit of
3^52?

(1) By far the simplest method: note the
following;

`3^0=1,3^1=3,3^2=9,3^3=27,3^4=81,3^5=243,3^6=729`

It
appears as though the last digit is following a pattern:1,3,9,7,1,3,9,... as indeed it
is. So every 4th power of 3 starting at 0 ends in a 1, so `3^52=(3^13)^4` ends in
1.

** This could be proved by induction if
needed**

(2) Consider that `3^52=(3^13)^4` . Now a typical
scientific calculator can evaluate `3^13` as 1594323. Now look at `(1594320+3)^4` . We
can use the binomial expansion theorem to
evaluate:

`(1594320+3)^4=`

`1594320^4+4(1594320)^3(3)+6(1594320)^2(9)+4(1594320)(27)+81`

Note
that each power of 1594320 ends in 0, so only the 81 contributes to the last
digit.

(3) `3^52=3^10*3^10*3^10*3^10*3^10*3^2` Again, a
typical scientific calculator can evaluate `3^10=59049` . So this product can be
written

(59040+9)(59040+9)(59040+9)(59040+9)(59040+9)(0+9).
Again all terms end in 0 except the term formed by the 9's, and
`9^6=531441`

(4) Imagine taking the 5 5-digit numbers from
(3) and the 9 and multiplying by hand. The only thing that affects the one's digit is
the product of those 6 nines.

(5) Finally, use a computer
program like Mathematica.

The last digit is a
1.

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