16¹⁶ = 6 (orice numar care se termina in 6 ridicat la orice putere se termina in 6)
32²³ = 2⁵⁾²³ = 2¹¹⁵
puterile lui 2 se repeta la interval de 4 (2, 4, 8, 6)
115 : 4 = 28 rest 3
32²³ = u(8) (cifra de pe pozitia a3a)
4⁸⁴ = 2²⁾⁸⁴ = 2¹⁶⁸
168 : 4 = 42 rest 0
4⁸⁴ = u(6)
64⁴⁶ = 2⁶⁾⁴⁶ = 2²⁷⁶
276 : 4 = 69 rest 0
64⁴⁶ = u(6)
n = 16¹⁶ + 32²³ + 4⁸⁴ + 64⁴⁶ = u(6) + u(8) + u(6) + u(6) = 26 = u(6)
