last digit problem: What are the last two digits of 2017^2017 ?

Firas

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What are the last two digits of 2017^2017?

Please provide a good method to find the last 2 digits please.
 
What are the last two digits of 2017^2017?

Please provide a good method to find the last 2 digits please.
Hint:

Start with =: what are the unit digits of 7^1, 7^2, 7^3, 7^4,7^5, 7^6 .... do you see a pattern?
 
that is the unit digit [correct]?
Yes, the last digit is in the units place. You need to start looking for patterns somewhere. Subhotosh's hint suggests looking for a pattern in the last digit, to start. Does the pattern give you any ideas, to continue?
 
Reply

can you please use modular arithmetic but explain as if you are explaining to a beginner because i have no idea how to use modular arithmetic to solve this!!! thanks.

because i cannot use a calculator in the actual exam so yeah.
 
can you please use modular arithmetic but explain as if you are explaining to a beginner because i have no idea how to use modular arithmetic to solve this!!! thanks.

because i cannot use a calculator in the actual exam so yeah.

Presumably you do know what modular arithmetic is, since you ask. Was that the context of the problem? It's a really good idea to state the context and what methods you are learning, if you want to get the most appropriate help.

To start with, we're going to look only for the last digit, not two digits; you're learning a new technique, so it's best to start small. That's the point of the hint you were given, and if you had followed it, you might have found the solution by now!

Here we go:

The last digit is congruent to the whole number, mod 10. (Do you understand that?)

So just do the arithmetic mod 10, meaning that you pay attention only to the last digit:

2017^2017 = 7^2017 (mod 10) -- do you see why?

Now start as was suggested, looking for a pattern, again doing the work in mod 10:

7^1 = 7
7^2 = 49 = 9 (mod 10)
7^3 = 7^2 * 7 = 9 * 7 = 63 = 3 (mod 10)

and so on. (It's actually easier just to think about what happens to the last digit as you repeatedly multiply by 7, without worrying about mod 10 explicitly!)

When you find a power of 7 that is congruent to 1, you will be able to use that result to jump ahead to the answer.

Now, that was just practice!

Do the equivalent in mod 100 next, in order to solve the actual problem.

For more information and ideas that haven't been mentioned yet, see Wikipedia, or search for the term.
 
Presumably you do know what modular arithmetic is, since you ask. Was that the context of the problem? It's a really good idea to state the context and what methods you are learning, if you want to get the most appropriate help.

To start with, we're going to look only for the last digit, not two digits; you're learning a new technique, so it's best to start small. That's the point of the hint you were given, and if you had followed it, you might have found the solution by now!

Here we go:

The last digit is congruent to the whole number, mod 10. (Do you understand that?)

So just do the arithmetic mod 10, meaning that you pay attention only to the last digit:

2017^2017 = 7^2017 (mod 10) -- do you see why?

Now start as was suggested, looking for a pattern, again doing the work in mod 10:



7^1 = 7
7^2 = 49 = 9 (mod 10)
7^3 = 7^2 * 7 = 9 * 7 = 63 = 3 (mod 10)

and so on. (It's actually easier just to think about what happens to the last digit as you repeatedly multiply by 7, without worrying about mod 10 explicitly!)

When you find a power of 7 that is congruent to 1, you will be able to use that result to jump ahead to the answer.

Now, that was just practice!

Do the equivalent in mod 100 next, in order to solve the actual problem.

For more information and ideas that haven't been mentioned yet, see Wikipedia, or search for the term.


Ok let me try :

last digit : 7 ^2017 (mod 10) : last digit is 2017 / 4 = .. R3 so last digit is 3
2nd last digit :

17^2017 (mod 100) = 17 / ??????
what do i do next?
 
Ok let me try :

last digit : 7 ^2017 (mod 10) : last digit is 2017 / 4 = .. R3 so last digit is 3
2nd last digit :

17^2017 (mod 100) = 17 / ??????
what do i do next?

also so last 3 digits means mod 1000?
 
Ok let me try :

last digit : 7 ^2017 (mod 10) : last digit is 2017 / 4 = .. R3 so last digit is 3
2nd last digit :

17^2017 (mod 100) = 17 / ??????
what do i do next?

Good, you've seen that 7^4 = 1 (mod 10); but 2017 does not leave remainder 3 on division by 4! Rather, the remainder is 1 (i.e. 2017 = 4*504 + 1). So 7^2017 = (7^4)^504 * 7^1 = 7. So the last digit is 7.

Now repeat the same basic idea. List powers of 17 (mod 100), and focus on those that end in 7. Look for a similar pattern. (There's a shortcut to avoid listing all the powers of 17; keep thinking as you work!)

Alternatively, you could look for a power of 17 that ends on 01, by doing something similar, and then use that power much as we did the power of 7.

also so last 3 digits means mod 1000?

Yes.
 
Good, you've seen that 7^4 = 1 (mod 10); but 2017 does not leave remainder 3 on division by 4! Rather, the remainder is 1 (i.e. 2017 = 4*504 + 1). So 7^2017 = (7^4)^504 * 7^1 = 7. So the last digit is 7.

Now repeat the same basic idea. List powers of 17 (mod 100), and focus on those that end in 7. Look for a similar pattern. (There's a shortcut to avoid listing all the powers of 17; keep thinking as you work!)

Alternatively, you could look for a power of 17 that ends on 01, by doing something similar, and then use that power much as we did the power of 7.



Yes.

thanks. i think it goes like this :

17^2017 = (17^21)*96 x 17
= ends with 1 x17
= 17

does this mean last 2 digits are 77. and last 3 are 177?

if that is correct, i hope you can help give me another similar question to test my understanding. Thanks so much!
 
thanks. i think it goes like this :

17^2017 = (17^21)*96 x 17
= ends with 1 x17
= 17

does this mean last 2 digits are 77. and last 3 are 177?

if that is correct, i hope you can help give me another similar question to test my understanding. Thanks so much!

i think i dont exactly get it yet. can you please explain (you dont need to give workings) how to get it from there. this isnt homework, I'm preparing for a test so I need understanding so I'm not going to just copy your answers. If you explain, I can understand better and solve other questions. like this : [from quora]. my answer was wrong as the answer is 56.




[FONT=q_serif]Using Modular Arithmetic,[/FONT]


[FONT=q_serif]last digit of 2016^2017 = 6^2017 (mod 10)

Therefore, the last digit is 6 as all powers of 6 end with 6.

last 2 digits of 2016^2017 = 16^2017 (mod 100)

16^2017 = (16^504)^4 x 16= (16^252)^8 x 16 = (16^126)^16 x 16= (16^63)^32 x 16= (16^21)^96 x 16

Therefore, (…12) x16 = 192

[/FONT]​
[/FONT]

 
thanks. i think it goes like this :

17^2017 = (17^21)*96 x 17
= ends with 1 x17
= 17

does this mean last 2 digits are 77. and last 3 are 177?

if that is correct, i hope you can help give me another similar question to test my understanding. Thanks so much!

I don't follow your reasoning, though I know that the last three digits are indeed 177.

Why did you use 17^21, whose last digits are 17, not 01? How do you conclude that the last digits are 77, having just arrived (incorrectly) at 17?

And why do you think you can get the last three digits without using mod 1000?

And you can easily make up a problem like this! The relatively hard part is to check your answer; for that, just go to wolframalpha.com and enter something like "2017^2017 mod 1000". You'll see if your answer is right.
 
I don't follow your reasoning, though I know that the last three digits are indeed 177.

Why did you use 17^21, whose last digits are 17, not 01? How do you conclude that the last digits are 77, having just arrived (incorrectly) at 17?

And why do you think you can get the last three digits without using mod 1000?

And you can easily make up a problem like this! The relatively hard part is to check your answer; for that, just go to wolframalpha.com and enter something like "2017^2017 mod 1000". You'll see if your answer is right.

thanks for that info. i understand i made errors. i get u have to use modular arithmetic. but once you get to lets say 17^9, how do you figure out the last 2 digits without listing out because i cant use a calculator?
 
thanks for that info. i understand i made errors. i get u have to use modular arithmetic. but once you get to lets say 17^9, how do you figure out the last 2 digits without listing out because i cant use a calculator?

Here is the first way I solved it (all work being mod 100):

We know that 7^4 = 1 (mod 10); so we look at 17^4, which is 21 (mod 100). We want a power that is equal to 1, so we try powers of 21 and find that 21^5 = 1. (All the work was done mod 100, so it is not hard -- you never need to write down digits to the left.)

So we now know that (17^4)^5 = 21^5 = 1, and 17^20 = 1. Since 2017 = 100*20 + 17, we have

2017^2017 = 17^2017 = 17^(20*100 + 17) = (17^20)^100 * 17^17 = 1^100 * 17^(4*4 + 1) = 1 * (17^4)^4 * 17^1 = 21^4 * 17 = 81 * 17 = 77.

There are other ways to do it, some perhaps easier. Work through this bit by bit to see why and how I did each step; then, if you want, look for a quicker way! Note, as I said, that you might start out just listing powers, but discover shortcuts as you do; this is a good way to learn. You don't need to know the best way from the start (though if you know questions like this will be on a test, you want to know at least a very good way before then).
 
Here is the first way I solved it (all work being mod 100):

We know that 7^4 = 1 (mod 10); so we look at 17^4, which is 21 (mod 100). We want a power that is equal to 1, so we try powers of 21 and find that 21^5 = 1. (All the work was done mod 100, so it is not hard -- you never need to write down digits to the left.)

So we now know that (17^4)^5 = 21^5 = 1, and 17^20 = 1. Since 2017 = 100*20 + 17, we have

2017^2017 = 17^2017 = 17^(20*100 + 17) = (17^20)^100 * 17^17 = 1^100 * 17^(4*4 + 1) = 1 * (17^4)^4 * 17^1 = 21^4 * 17 = 81 * 17 = 77.

There are other ways to do it, some perhaps easier. Work through this bit by bit to see why and how I did each step; then, if you want, look for a quicker way! Note, as I said, that you might start out just listing powers, but discover shortcuts as you do; this is a good way to learn. You don't need to know the best way from the start (though if you know questions like this will be on a test, you want to know at least a very good way before then).

ok, but how did you know 17^4 is 21? how did you use mod 100 to get that? thanks
 
ok, but how did you know 17^4 is 21? how did you use mod 100 to get that? thanks

17^2 = 17*17 = 289 = 89
17^3 = 89*17 = 1513 = 13
17^4 = 13*17 = 221 = 21

Just keep multiplying, retaining only the last two digits.

If you are doing it on paper, you don't even have to write anything but those two digits if you don't want to:

Code:
[FONT=courier new]  89
* 17
----
 *23
*89
----
**13[/FONT]

If you think in mod 100 at every step, the work is kept as easy as possible.
 
17^2 = 17*17 = 289 = 89
17^3 = 89*17 = 1513 = 13
17^4 = 13*17 = 221 = 21

Just keep multiplying, retaining only the last two digits.

If you are doing it on paper, you don't even have to write anything but those two digits if you don't want to:

Code:
[FONT=courier new]  89
* 17
----
 *23
*89
----
**13[/FONT]


If you think in mod 100 at every step, the work is kept as easy as possible.

oh! so the third digit doesnt really matter? thank you!
 
Here is the first way I solved it (all work being mod 100):

We know that 7^4 = 1 (mod 10); so we look at 17^4, which is 21 (mod 100). We want a power that is equal to 1, so we try powers of 21 and find that 21^5 = 1. (All the work was done mod 100, so it is not hard -- you never need to write down digits to the left.)

So we now know that (17^4)^5 = 21^5 = 1, and 17^20 = 1. Since 2017 = 100*20 + 17, we have

2017^2017 = 17^2017 = 17^(20*100 + 17) = (17^20)^100 * 17^17 = 1^100 * 17^(4*4 + 1) = 1 * (17^4)^4 * 17^1 = 21^4 * 17 = 81 * 17 = 77.

There are other ways to do it, some perhaps easier. Work through this bit by bit to see why and how I did each step; then, if you want, look for a quicker way! Note, as I said, that you might start out just listing powers, but discover shortcuts as you do; this is a good way to learn. You don't need to know the best way from the start (though if you know questions like this will be on a test, you want to know at least a very good way before then).

i recently have seen someone using binomial theorem to expand 17^2017. how is that possible?
 
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