Assuming that you have $N$ ciphertexts with different keys each, finding one of the plaintext becomes easier as $N$ increases ($2^{128 - N + k}$ tries - where $k$ is $2^k$ = the amount of plaintexts you want to find), making it actually feasible to find some plaintexts of AES-128 with a large amount of ciphertexts in a reasonable time.

Also see https://cr.yp.to/snuffle/bruteforce-20050425.pdf or if you want a simpler explanation https://blog.cr.yp.to/20151120-batchattacks.html

Assuming that you have $2^N$ ciphertexts with different keys each, finding one of the plaintext becomes easier as $N$ increases ($2^{128 - N + k}$ tries - where $k$ is $2^k$ = the amount of plaintexts you want to find), making it actually feasible to find some plaintexts of AES-128 with a large amount of ciphertexts in a reasonable time.

Also see https://cr.yp.to/snuffle/bruteforce-20050425.pdf or if you want a simpler explanation https://blog.cr.yp.to/20151120-batchattacks.html

Assuming that you have N ciphertexts with different keys each, finding one of the plaintext becomes easier as N increases (2^(128 - N + k) tries - where k is 2^k = the amount of plaintexts you want to find), making it actually feasible to find some plaintexts of AES-128 with a large amount of ciphertexts in a reasonable time.

Also see https://cr.yp.to/snuffle/bruteforce-20050425.pdf or if you want a simpler explanation https://blog.cr.yp.to/20151120-batchattacks.html

Assuming that you have $N$ ciphertexts with different keys each, finding one of the plaintext becomes easier as $N$ increases ($2^{128 - N + k}$ tries - where $k$ is $2^k$ = the amount of plaintexts you want to find), making it actually feasible to find some plaintexts of AES-128 with a large amount of ciphertexts in a reasonable time.

Also see https://cr.yp.to/snuffle/bruteforce-20050425.pdf or if you want a simpler explanation https://blog.cr.yp.to/20151120-batchattacks.html