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Using modular arithmetic would be the most elegant and fastest way to determine the answer. So,

First write $16^{123}$ as $16 \times (16^2)^{61}$

Now, we just need to use the property that $16^2 \equiv -1 \pmod{257} \Rightarrow (16^2)^{61} \equiv (-1)^{61} \equiv -1 \pmod{257}$

So, $16 \times (16^2)^{61} \equiv -16 \equiv 241 \pmod{257}$

We start by working out a few examples. $16^1$ has a remainder of 16 when divided by 257. $16^2$ has a remainder of 256 when divided by 257. Using modular arithmetic, the remainder can be expressed as -1. $16^3$ has a remainder of 241 when divided by 257. The remainder can similarly be expressed as -16. At this point, a pattern seems to emerge, so we continue. $16^4$ has a remainder of 1 and $16^5$ has a remainder of 16.

Thus the pattern of remainders when $16^x$ is divided by 257 is $16,-1,-16,1$ We can use this pattern to our advantage and see where the power 123 lands. $123/4$ has a remainder of 3, so it lands on -16.

$16^{123}$ has a remainder of -16, or 241.

16^{ 123 }

=(257-1)^{ 123 } X 16

When (257-1)^{123} is divided by 257, remainder will be -1.

When 16 is divided by 257, remainder will be 16.

-1 X 16 = -16.

To get remainder, subtract 16 from 257.

                                 257 - 16 = 241

We can use the fact that $256 == 16^{2}$ , then we can transform $16^{123}$ to $16 * 256^{61}$ , and since

$256 \times 256 mod 257 == 1$ , then $16 \times 256^{61} mod 257 === 16 \times 256 mod 257 == 241$

16^2=-1(mod 257). Raising both sides to the 61st power: 16^122=-1(mod 257). Multiplying both sides by 16: 16^123=-16(mod 257). Therefore, the remainder when 16^123 is divided by 257 is 241.

First $16^{123}$ = $2^{492}$ and 257 = $2^{8}$ -1. Let's do the division of the polynomial $x^{492}$ for $x^{8}$ -1. The quotient is ( $x^{484}$ - $x^{476}$ + $x^{468}$ +...+ $(-1)^{k}$ $x^{484-8k}$ ) for k=1..60 and the remainder is $(-1)^{k+1}$ $x^{484-8k}$ for k = 60. (It derive from the algorithm of the divion of polynomial) So for x=2 we have $2^{492}$ = ( $2^{484}$ - $2^{476}$ + $2^{468}$ +... $-2^{12}$ + $2^{4}$ ) - $2^{4}$ . Then, the reminder is - $2^{4}$ = -16 and so the reminder is 257-16 = $\boxed {241}$

Using Modular Arithmetic:

16^{2} = -1 (mod 257)

Raise -1 to the power of 60 to find 16^{120} (mod 257)

=1 (mod 257)

16^122 = 1 \times 16 ^{2} (mod 257)

=-1

16^{123}=-1 /times 16

=-16 (mod 257) = 241 (mod 257)

Given $[p\cdot q]_a = [[p]_a \cdot [q]_a]_a$ and $[16^4]_{257} = 1$

$[16^{123}]_{257} = [16^{4 \cdot 30 + 3}]_{257} = [16^{4 \cdot 30} \cdot 16^3]_{257}$

$= [([16^4]_{257})^{30} \cdot [16^3]_{257}]_{257}$

$= [1^{30} \cdot [16^3]_{257}]_{257}$

$= 1 \cdot [16^3]_{257}$

$= \boxed{241}$

16%257=16

16^2%257=256

16^3%257=241

16^4%257=1

16^5%257=16 and so we find a sequence here, the modulus repeats as a sequence of 4 alternate steps

now, 123%4=3.

and, we see from the above that for power of 3 , the remainder is 241.

We know that

$16^{123} = 16^{100} \times 16^{20} \times 16^{3}$

So $\frac{16^{123}}{257} = \frac{16^{100} \times 16^{20} \times 16^{3}}{257}$

so by

$\frac{16^{3}}{257}$

= $\frac{4096}{257}$

Therefore, We get the remainder $\boxed{241}$

Consider that ­ $257 ­= 256 + 1$ and $16^{123} = 16 \times 256^{61}$ . Since $256^{n} \mbox { (mod } 257) \equiv -1$ for all odd integers $n$ and $61$ is odd, $16 \times 256^{61} \mbox { (mod } 257 ) \equiv 16 \times -1 \equiv -16 \equiv 241 \mbox { (mod } 257 )$

It's obvious, isn't it? My method is by trial. GG bros