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The value of $3^{51} \text{ mod } 5$ is _____
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Answer : 2

cycle of 3 where unit place repeat is 4 . and unit digit of last cycle is 1 . 

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

now repeat

3^5 = 3

so, (3^4)^12 * 3^3 mod(5) = 1*3^3 mod(5) = 27 mod(5) = 2

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This can be done by using Euler Totient function

Φ(5) = 4(as 5 is a prime number all thwe positive numbers less than 5 are co-prime to it)

now 51%4 = 3

$3^3 \%5 = 27\%5 = 2$

 

so answer is 2
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$3^1$=3                         $3^2$ =9                      $3^3$=27                  $3^4$=81           (Last digits are 3,9,7,1)

$3^5$=243                    $3^6$=729                  $3^7$=2187             $3^8$=6561        (Last digits are 3,9,7,1)

$3^9$=19683    and so on…

Observe that for 3 power’s interval of 4, the last digits repeat from 3.

So given power of 3 is 51:  4*12=48. Now from 3’s power 49, the last digit will start from 3.

$3^{49}$ =last digit is 3

$3^{50}$ =last digit is 9

$3^{51}$ =last digit is 7

So $3^{51}$=______7 mod 5 = 2.
Answer:

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