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

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## 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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We will note the sequence of the ending digit:

3^1 = **3**

3^2 = **9**

3^3 = 2**7**

3^4= 8**1**

3^5= 24**3**

3^6= 72**9**

3^7 = 218**7**

Looking at the first digit of all numbers we notice the pattern 3, 9, 7, 1, 3, 9, 7, 1 ,....

The powers 1, 5, 9, 13, 17, ....has ending digit 3

The powers 2, 6, 10, 14, ... ...has the ending digit 9

The powers 3, 7, 11, 15, .... has the ending digit 7

The powers 4, 8, 12, 16, .... has the ending digit 1

Now let us analyze the power 52.

52 = 2*2*13 = **4***13

Then it belongs to the 4th sequence which is ending with the digit 1.

**Then, the last digit of the number 3^52 is 1.**