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.
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.
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.
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...
Well, I'll try.
So, we're repeatedly multiplying by 3; note what happens when we multiply some multi-digit number by 3:
Notice that the 8 and 0 in 802 have no effect on the last digit of the product; only the 2 matters.Code:802 x 3 ------ 2406
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
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 toCode: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 ...
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.
Hello, zunecrazy!
You sound like you don't know what a "one's digit" is . . .
It is the rightmost (last) digit.
We are concerned with the last digit only.
Consider consecutive powers of 3 . . . and their last digits.
. .
We see that ends in 1.
Note that: .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . This ends in 1