SOLUTION: What is the remainder when 23^294 is divided by 5? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4

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Question 883926: What is the remainder when 23^294 is divided by 5?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Focus only on the last units digit of 23

3 ---> 3^2 = 9

9*3 = 27

Now only focus on the 7 and multiply by 3: 7*3 = 21

Now only focus on the 1 and multiply by 3: 1*3 = 3

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We have this pattern of the digits: 3, 9, 7, 1

and it repeats over and over

It repeats every 4 times. So because 294/4 = 73 remainder 2, this means that the last digit of 23^294 is 9

9/5 = 1 remainder 4

So the remainder is 4


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
What is the remainder when 23^294 is divided by 5?
294 = 2^8 + 2^5 + 2^2 + 2
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23 = 3 (mod 5)
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So 23^294 Mod 5 = 23^(2+2^2+2^5+2^8) mod5
= 3^2*3^4*3^5*3*8 (mod5)
= 4*1*2*3 (mod5)
= 4 (mod5)
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Cheers,
Stan H.
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(A) 0
(B) 1
(C) 2
(D) 3
(E) 4