See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Here,  f n  (sum of digits of natural number n)2      2 f 2011 2 0 1 1 16               2 2f 2011 f f 2011 f 16 1 6 49      For More Material Join: @JEEAdvanced_2024 AITS-CRT-II (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 13       2 3f 2011 f 49 4 9 169           2 4f 2011 f 169 1 6 9 256            2 5f 2011 f 256 2 5 6 169            2 6f 2011 f 169 1 6 9 256         2n f 2011 256  and   2n 1 f 2011 169           2017 2016 2017 2018 f 2011 f 2011 169 256 1 169 256 f 2011 f 2011       