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1) Find the remainder of 5^400 when divided by 19

 

2) What is the tens digit of 7^2018

 

3) Solve the congruence: 3x = 5(mod 4), k is any integer

 

 

TYSM :)

 Mar 14, 2021
 #1
avatar+14 
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2) 7^1 ones digit=7

7^2 ones digit=9

7^3 ones digit=3

7^4 ones digit=1

7^5 ones digit=7

loop of 4, 2018 is congruent to 2(mod 4)

7^2018 ones digit=9

(I think, I'm only in 7th grade and I'm new to this,so I'm not as smart as CPhill, so try to get a second opinion)

 Mar 14, 2021
edited by Guest  Mar 14, 2021
edited by fuhsaha  Mar 14, 2021
 #2
avatar+291 
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Fuhsaha, I think that's correct, but the question asked for the tens digit.

1) Since 19 is prime, Fermat's Little Theorem is applicable:

\(5^{400} = 5^{22(18)+4}\\ 5^{22(18)+4} {\displaystyle \equiv } 5^4 (\text{mod 19})\\5^4 = 625\\625{\displaystyle \equiv }17(\text{mod 19})\)

This means the remainder is \(\boxed{17}\)

2) I know there is a way of doing this using pure modular arithmetic by taking that number mod 100, but I still don't really know how to do it. Here's the look-for-cycle method just like fuhsaha used:

7^1 = 07 (last 2 digits)

7^2 = 49 

\(\)7^3 = 43

7^4 = 01

7^5 = 07 

(by the way, you don't actually need to compute 7^3, 7^4, or 7^5, you just need to multiply the last 2 digits of the previous power by 7.)

That means there is a cycle of 4, and the last 2 digits of 7^2018 is 49, which means that the 10's digit is \(\boxed{4}\)

3) where is k? 

 Mar 14, 2021
 #3
avatar+130 
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Thank you very much!

HelpPls123abc  Mar 15, 2021

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