First, we see that $1993 \equiv 4 \pmod{13}$ .

Therefore, the remainder when $1993$ is divided by $13$ will be the same when $4$ is divided by $13$ .

Therefore, we need to find the $n$ such that $4^{1993} \equiv n \pmod{13}$ .

Then, we see that $4^{3} \equiv - 1 \pmod{13}$ . Lets call this result important result.

Now, let´s write $4^{1993}$ as $(4^{3})^{664} \times 4$ .

That means that $(4^{3})^{664} \times 4 \equiv n \pmod{13}$ .

Then, by the important result (see above), $(- 1)^{664} \times 4 \equiv n \pmod{13}$ .

$\Rightarrow 1 \times 4 \equiv n \pmod{13}$

$\Rightarrow 4 \equiv n \pmod{13}$

$\boxed{1993^{1993} \equiv 4 \pmod{13}}$

We observe that 1993 is congruent to 4 (mod 13). So it's essentially asking what is the remainder when 4^1993 is divided by 13? Now 4^6 is congruent to 1 (mod13). So 4 ^(6*332) = 4^1992 is congruent to 1(mod13). Multiplying both sides by 4 we get 4^1993 is congruent to 4(mod 13). Hence the remainder is 4.

the remainder of 1993/13 is 4. So it like 17/13. 17^1/13 remain 4, 17^2/13 remain 3, 17^3/13 remain 12, 17^4/13 remain 9, 17^5/13 remain 10, 17^6/13 remain 1, 17^7/13 remain 4, 17^8/13 remain 3 and so on... So the loop of the remainder is 6. And 1993/6 remain 1 <=> 1993^1993 remain 4

By Fermat's Little theorem we have 1993^12 congruent to 1 modulo 13 Rasing power by 166 we get 1993^1992 congruent to 1 modulo 13 1993 is congruent to 4 modulo 13 By multiplying these congruence relations we get 1993^1993 congruent to 4 modulo 13. Hence, 4 is the remainder.

Observe that $\displaystyle 1993^{1993} \equiv 4^{1993} \bmod{13}$ .

Since $gcd(4, 13) = 1$ , then, by Euler's theorem , $4^{\phi (13)} \equiv 4^6 \equiv 1 \bmod{13}$ .

But $4^{1993} \equiv 4(4)^{1992} \equiv 4{(4^{6})}^{332} \equiv 4(1)^{332} \equiv 4 \bmod{13}$ .

Therefore $\displaystyle 1993^{1993} \equiv 4 \bmod{13}$ .

Fermat's little theorem states that: If $p$ is prime and $m$ and $n$ are positive integers such that:

$m\equiv n\pmod{p-1}$ then for every integer $a$ we have $a^m\equiv a^n\pmod{p}$ .

Since $13$ is a prime and $m = 1993 \equiv 1 \pmod{12}$ , then:

$1993^{1993} \equiv 1993^1 \equiv (153\times 13 + 4) \equiv \boxed {4} \pmod {13}$

just simply divide 1993 to 13

the remainder is 4

First I divided 1993 with 13 and got remainder 4 then squared and done same thing then got 3 then cube got 2 then power 4 got 1 .so I subtracted how many 4's are there in power 1993 so only power 1 is left .t first I got 4 by dividing 1993 with13 That's the answer

See that $1993 \equiv 4 \mod 13$

So we need to find the remainder of dividing $4 ^ {1993}$ by $13$ Now we need to find the exponent such that $4 ^ n \equiv 1 \mod 13$ This exponent is $6$ .

$4 ^ 6 \equiv 1 \mod 13$

Note that $4^{1993} =(4^{6})^{332} \cdot 4$

$(4 ^ 6) ^{332} \equiv 1^{332} \mod 13\\ 4 ^ {1992} = 1 \mod 13\\ 4 ^ {1992}\cdot 4 \equiv 1 \cdot 4 \mod 13\\ 4 ^ {1993} \equiv 4 \mod 13$

The remainder is 4

1993 / 13 => remainder = 4 hence for 1993 ^ n => remainder = 4

1983 is congruent to 4 mod 13, 1993^2 is congruent to 3 mod 13, (1993^2)^3 is congruent to 1 mod 13, (1993^6)^332 is congruent to 1 mod 13, therefore 1993^1992 is congruent to 1 mod 13, So (1993^1992)x1993 is congruent to 1993 mod 13 , or, 1993^1993 is congruent to 4 mod 13.

We can simplify this problem by taking the base 13 modulo of 1993 and raising it to the appropriate power: $1993^{1993} \equiv 4^{1993} \pmod{13}$ We can now rearrange the powers so that we can use this method once again: $4^{1993}=4 \times (4^{2})^{996} \equiv 4 \times 3^{996} \pmod{13}$ We can rearrange once more: $4 \times 3^{996} = 4 \times (3^{12})^{83} \equiv 4 \times 1^{83} \pmod{13}$ We thus conclude that: $1993^{1993} \equiv \boxed{4} \pmod{13}$

I wish you understand my solution.

The problem above can be expressed as $1993^{1993} \pmod{13}$ .

You can find the value of $1993 \pmod{13}$ first to make the problem more bearable. By the division algorithm, we can see that the remainder when $1993$ is divided by $13$ is $4$ .

$1993 = 153(13) + 4$

Now, we can replace $1993$ by $4$ being equivalent.

$1993^{1993} \pmod{13}$

$\equiv 4^{1993} \pmod{13}$

Looking into the powers of $4$ , $4^3 \pmod{13} = 64 \pmod{13} \equiv 12$ because $64 = 4(13) + 12$ . You will see why I chose $4^3$ among all the possible values.

$\equiv (4^3)^{664} \times 4 \pmod{13}$

$\equiv (12)^{664} \times 4 \pmod{13}$

Look at this: Since $12^2 = 144$ and $13 \times 11 = 143$ , then $12^2 \pmod{13} = 144 \pmod{13} \equiv 1$ . Therefore,

$\equiv (12^2)^{332} \times 4 \pmod{13}$

$\equiv 1^{332} \times 4 \pmod{13}$

$\equiv 4 \pmod{13}$

$\equiv \boxed{4}$

Apply Fermat's little theorem for the problem, It states that if P is a prime number and a is any integer such that a and P are relatively prime,then a^(P-1)= 1 mod P. In this case a=1993,P=13 , 1993^(12)=1 mod 13, 1993^(1992)=1^1992 mod 13=1 mod 13, 1993^(1993)=1993 mod 13 =4 mod 13. Hence remainder of 1993^(1993) when divided by 13 is 4.

We have $1993 \equiv 4 \pmod {13}$ , so $4^{1993}=4\cdot 4^{1992}=4 \cdot (4^3)^{664} \equiv 4$ , because $4^3=64 \equiv -1 \pmod {13}$ .

here's a c code ans: 4

include<stdio.h>

int main() { int k, l, sum=1; for(k=0; k<1993; k++) { sum=sum*1993; } printf("%d", sum%13); }

We can change the problem above to 1993 = 153.13 + 4 ----> 4^1993 mod 13 = 4.(4^4)^498 mod 13 = 4.(256)^498 mod 13 = 4(19.13 + 9)^498 mod 13 = 4(9)^498 mod 13 = 4(3^2)^498 mod 13 = 4.(3^4)^249 mod 13 = 4.(81^249) mod 13 = 4.(13.6 + 3)^249 mod 13 = 4(3^3)^83 mod 13 = 4(27)^83 mod 13 = 4(2.13 + 1)^83 mod 13 = 4.(1^83) mod 13 = 4.1 mod 13 = 4 mod 13. Answer : 14

Note that the remainder when $1993$ is divided by $13$ is $4$ . Then, we're looking for the remainder when $4^{1993}$ is divided by $13$ . The remainder of $\frac {4^{2}}{13}$ is $3$ . So this is the same than $3^{996} \cdot 4$ . The remainder of $\frac {3^{3}}{13}$ is $1$ . Hence, this is $1^{332} \cdot 4 = \boxed {4}$ .

First we simplify 1993 mod 13 to 4 mod 13. When dealing with remainders and modulo, it is ideal to look for a pattern (one is almost surely to come up). So we experiment with the first few powers of four and their remainder. We get the pattern

$4, 3, 12, 9, 10, 1, 4, 3, 12 ...$

We see that every six terms, the pattern repeats. To find the term, we look at the remainder of the power: 1993 / 6 = 332 R 1. Thus the answer is $\boxed{4}$