Surely it is largest?

$357^{2468}$ $\equiv x$ (mod 456)

456 = $3 \times8 \times19$

$STEP 1$

357 $\equiv 0$ (mod 3)

357 $\equiv 0$ (mod 3)

$357^{2468}$ $\equiv 0$ (mod 3)

$STEP 2$

357 $\equiv 5$ (mod 8)

$357^{2}$ $\equiv 1$ (mod 8)

$357^{2(1234)}$ $\equiv 1$ (mod 8)

$357^{2468}$ $\equiv 1$ (mod 8)

$STEP 3$

357 $\equiv 15$ (mod 19)

$357^{18}$ $\equiv 1$ (mod 19) (according to Fermat's little theorem)

$357^{18(137)}$ $\equiv 1$ (mod 19)

$357^{18(137)}$ X $357^{2}$ $\equiv 225$ (mod 19)

$357^{2468}$ $\equiv 225$ (mod 19)

$STEP 4$

Therefore,

$357^{2468}$ leaves a remainder as

0 when divided by 3

1 when divided by 8

225 (or) 16 when divided by 19

If we consider 225 as x, it is suiting our requirements!

Thus, the answer is $\huge{225}$