# Thread: This Question Concerns the Number 3 to the 2,719th power

1. ## This Question Concerns the Number 3 to the 2,719th power

Hi there. I got this problem in my liberal arts math class. It has me stumped. It goes like this.

This Question Concerns the Number 3 to the 2,719th Power.

1) What is the one's digit of this number? Show/Explain your work.

2) In arriving at your answer to #1, did you rely on deductive or inductive reasoning? Explain.

I guess my main question is how to even find the one's digit.

Any help is very appreciated.

2. Originally Posted by zunecrazy
Hi there. I got this probem in my liberal arts math class. It has me stumped. It goes like this.

This Question Concerns the Number 3 to the 2,719th Power.

1) What is the one's digit of this number? Show/Explain your work.

2) In arriving at your answer to #1, did you rely on deductive or inductive reasoning? Explain.

I guess my main question is how to even find the one's digit.

Any help is very appreciated.
You use arithmetic modulo 10, and build up to the answer, so:

$\displaystyle 3^2=9 {\text{ mod }} 10$
$\displaystyle 3^4=1 {\text{ mod }} 10$
$\displaystyle 3^8=1 {\text{ mod }} 10$
:
:
etc

From these observe that (and check it, arithmetic is not guaranteed):

$\displaystyle 3^{2719}=3^{2^{11}}3^{2^9}3^{2^7}3^{2^4}3^{2^3}3^{ 2^2}3^23$

CB

3. Thanks. Although I'm still pretty confused. Is that done on a calculator? I'm pretty terrible with math.

4. Consider the first several powers of 3 and their last digits: 1 (= 3^0), 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, etc. Then note that the last digit of 3n is only determined by the last digit of n, for any n. One can see this from long multiplication, for example. Or write n = 10m + k: then 3n = 10 x 3m + 3k, so m influences only tens and up.

5. Originally Posted by zunecrazy
Hi there. I got this problem in my liberal arts math class. It has me stumped. It goes like this.

This Question Concerns the Number 3 to the 2,719th Power.

1) What is the one's digit of this number? Show/Explain your work.

2) In arriving at your answer to #1, did you rely on deductive or inductive reasoning? Explain.

I guess my main question is how to even find the one's digit.

Any help is very appreciated.
You can notice that the ones digit of 3^n is cyclic with period 4.

3^1 = ...3
3^2 = ...9
3^3 = ...7
3^4 = ...1
3^5 = ...3

Thus, find the remainder of 2719 when divided by 4 and raise 3 to that, and look at the ones digit. If you continue with mathematics until number theory you'll recognise this later as Euler's theorem.

6. Thank you all. What do you think of this solution?

It is either 3 9 7 or 1.
3 to power 1 is 3
3 to power 2 is 9
3 to power 3 is 27
3 to power 4 is 81
3 to power 5 is 243
3 to power 6 is 729
3 to pwer 7 is 2187
3 to power 8 is 6561
so the sequence of unit column (1's column) repeats every 4 powers. Therefore If I count correctly, 2719th power would be a equivalent to power 9 and the UNIT/ONES number should be a "9"

Deductive reasoning...

7. Originally Posted by zunecrazy
Thank you all. What do you think of this solution?

It is either 3 9 7 or 1.
3 to power 1 is 3
3 to power 2 is 9
3 to power 3 is 27
3 to power 4 is 81
3 to power 5 is 243
3 to power 6 is 729
3 to pwer 7 is 2187
3 to power 8 is 6561
so the sequence of unit column (1's column) repeats every 4 powers. Therefore If I count correctly, 2719th power would be a equivalent to power 9 and the UNIT/ONES number should be a "9"

Deductive reasoning...
You should end up with 7.

Also you should justify why there is a cycle with period 4. emakarov provided justification above.

8. Maybe I'm being dense but I really really don't understand at all. I'm terrible at math. If anyone could break this down to the most simplest, layman terms I would be forever grateful.

9. Originally Posted by zunecrazy
Maybe I'm being dense but I really really don't understand at all. I'm terrible at math. If anyone could break this down to the most simplest, layman terms I would be forever grateful.
Well, I'll try.

So, we're repeatedly multiplying by 3; note what happens when we multiply some multi-digit number by 3:

Code:
   802
x    3
------
2406
Notice that the 8 and 0 in 802 have no effect on the last digit of the product; only the 2 matters.

So, 23984792387498237498237498327948327 * 3, what's the last digit? You can ignore all the digits except for the last, and immediately reply, the last digit is 1.

So,

3^0 ends with 1.
3^1 ends with 3.
3^2 ends with 9.
3^3 ends with 7.
3^4 ends with 1.

We are back where we started, and the cycle must continue.

Then notice this

Code:
n         r
------------
0 = 0*4 + 0
1 = 0*4 + 1
2 = 0*4 + 2
3 = 0*4 + 3
4 = 1*4 + 0
5 = 1*4 + 1
6 = 1*4 + 2
...
n represents the exponent, and r represents the remainder when n is divided by 4. Knowing r tells us what the last digit of 3^n is. We go according to

3^0 ends with 1.
3^1 ends with 3.
3^2 ends with 9.
3^3 ends with 7.
...

So for the actual problem, we want to know what the remainder of 2719 is when divided by 4. Turns out it is 3. Therefore the last digit of 3^2719 is 7.

10. Originally Posted by CaptainBlack
You use arithmetic modulo 10, and build up to the answer, so:

$\displaystyle 3^2=9 {\text{ mod }} 10$
$\displaystyle 3^4=1 {\text{ mod }} 10$
$\displaystyle 3^8=1 {\text{ mod }} 10$
:
:
etc

From these observe that (and check it, arithmetic is not guaranteed):

$\displaystyle 3^{2719}=3^{2^{11}}3^{2^9}3^{2^7}3^{2^4}3^{2^3}3^{ 2^2}3^23$

CB
So modulo 10 we have:

$\displaystyle 3^{2719} {\text{ mod }}10=1 \times 1 \times 1 \times 1 \times 1 \times 1\times 9 \times 3 {\text{ mod }}10=7 {\text{ mod }}10$

CB

11. Hello, zunecrazy!

You sound like you don't know what a "one's digit" is . . .
It is the rightmost (last) digit.

$\displaystyle \text{Determine the one's digit of: }\:3^{2719}$

We are concerned with the last digit only.

Consider consecutive powers of 3 . . . and their last digits.

. . $\displaystyle \begin{array}{cc} \text{Power} & \text{ends in:} \\ \hline 3^1 = 3 & 3 \\ 3^2=9 & 9 \\ 3^3=27 & 7 \\ 3^4=81 & 1 \\ \vdots & \vdots \end{array}$

We see that $\displaystyle 3^4$ ends in 1.

Note that: .$\displaystyle 2719 \:=\:(4)(679) + 3$

$\displaystyle \text{Hence: }\;3^{2719} \;=\;3^{4(679) + 3} \;=\;3^{4(679)}\cdot3^3 \;=\;(3^4)^{679}\cdot3^3$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \uparrow$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
This ends in 1

$\displaystyle \text{Therefore, }\,3^{2719}\,\text{ ends in }\,1^{679}\cdot27\quad\hdots\;\text{ which ends with 7.}$