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# help!

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two problems of the same type:

what is the remainder of 7^2019/5?

what is the remainder of 333^333 / 33?

thanks

Apr 28, 2019

#1
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1. Simple pattern recognizing.

Remainder of when divided by 5-

7^1=2

7^2=4

7^3=3

7^4=1

7^5=2

and so on

This means the remainder pattern is 2, 4, 3, 1,

To find the remainder of 7^2019, we find the remainder of 2019 divided by 4 because there are 4 terms in the pattern.

Since the remainder is 3, we count up 3 times in the pattern.

So the remainder of 7^2019 divided by 5 is 3.

Apr 28, 2019
#2
+1

Thanks! I wasn't sure if Modulo 7^2019 worked on this kind of problem, and yeah, I think it does.

So my way (that I didn't know that worked) was to see the ones digit for powers of the remainder of 7 mod 5.

2, 4, 8, 6

It would be 8, and then divide that by 5 and get the remainder of 3.

Thanks!

Starz  Apr 28, 2019
#3
+4

That is pretty ingenious and is better than what I did!

Apr 28, 2019
#4
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333^333 mod 33 = 27

333^1 mod 33 = 3

333^2 mod 33 = 9

333^3 mod 33 =27

.

.

.

333^8 mod 33 =27

333^333 mod 33 = 27

Apr 28, 2019
edited by Guest  Apr 28, 2019
edited by Guest  Apr 28, 2019
edited by Guest  Apr 28, 2019