What are the last two digits of 2017^{2017}?

N

nanase

Junior Member
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Aug 8, 2019
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I need help in working this out
last 2 digits.jpg
No other hint is given. What I tried is looking for pattern by typing it to the power of 1 , 2 and 3 and so on, but it gets too large and incomputable. What other way can I use and how? Thank you
 
BigBeachBanana said:
Hint: Modulo 100
Click to expand...
Hi BBB
I tried this which gives 17, then I search for pattern
like 17^1 , 17^2 and so on, but the last two digits don't form any distinguishable pattern. Also I am not supposed to use calculator for this? can you give more hints/help?
 
nanaseailie said:
Hi BBB
I tried this which gives 17, then I search for pattern
like 17^1 , 17^2 and so on, but the last two digits don't form any distinguishable pattern. Also I am not supposed to use calculator for this? can you give more hints/help?
Click to expand...
[math]17^{17} = 17^{16}\times 17[/math]
Can you figure out the following even powers?
[imath]17^{2},17^{4},17^{8},17^{16} \mod 100?[/imath]

Edit:
Use [imath]17^2 \mod 100[/imath] to figure out [imath]17^4 \mod 100[/imath] and so on...
 
Last edited:
The only thing I understand is the indices property applied there, but I am lost can you help further?
 
ans2.jpg
I can't continue because the calculator won't display the last 2 digits anymore.
but I am not supposed to use calculator so this is probably not the right method also. But what am I seeing here? I don't see any pattern yet
 
nanaseailie said:
View attachment 36679
I can't continue because the calculator won't display the last 2 digits anymore.
but I am not supposed to use calculator so this is probably not the right method also. But what am I seeing here? I don't see any pattern yet
Click to expand...
Read post#8 again. You're not looking for a pattern, but leveraging prior modulo to find the subsequent ones. We don't need all of the even powers, only 2,4,8,16. You shouldn't need a calculator, only multiply 2 digits by 2 digits.
 
I re-do the above working
ans3.jpg
please the me if this is correct, the last 2 digits are 77.
I use all the hints given above.
 
BigBeachBanana said:
The final answer is correct but your intermediate steps are not.
[imath](17^4)^2 \neq 17^6[/imath]
Click to expand...
ans3.1.jpeg
Thank you for correcting my steps, appreciate it as I am here not just for the answer, but for understanding how things work also mathematically
Thank you muchhhh Mr BBB.
Is the above correct ?
 
I suspect that you will NOT find a pattern with the last two digits of powers of 17. mod 100 will help.

Why are you looking at all even powers of 17?

Only consider 17, 17^2, 17^4, 17^8, ....
How do you go from one to the next?
 
nanaseailie said:
View attachment 36690
Thank you for correcting my steps, appreciate it as I am here not just for the answer, but for understanding how things work also mathematically
Thank you muchhhh Mr BBB.
Is the above correct ?
Click to expand...
Maybe I missing something but I don't think your answer is correct as I don't see your answer.
An answer should look like the last two digits in 2017^2017 is _ _ !!!!!!
 
Steven G said:
Maybe I missing something but I don't think your answer is correct as I don't see your answer.
An answer should look like the last two digits in 2017^2017 is _ _ !!!!!!
Click to expand...
in post #11 I put 77, Mr Banana is correcting me on my steps.
 
nanaseailie said:
in post #11 I put 77, Mr Banana is correcting me on my steps.
Click to expand...
No, I won't accept your reply--sorry
In post 13 you asked if the work in that post is correct.
If someone gives me some work to look at I would NEVER say that it is correct when it is incorrect just because the work they showed me before was correct.
 
Steven G said:
No, I won't accept your reply--sorry
In post 13 you asked if the work in that post is correct.
If someone gives me some work to look at I would NEVER say that it is correct when it is incorrect just because the work they showed me before was correct.
Click to expand...
noted, you have a point. Thank you so much for the advise :thumbup:
 
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