Solution: Compute the sum: $ 12003 + 12005 + 12007 + 12009 $. This is an arithmetic sequence with first term 12003, last term 12009, 4 terms. Sum = $ rac42 (12003 + 12009) = 2 \cdot 24012 = 48024 $. Now compute $ 48024 \mod 16 $. Note that $ 10000 \equiv 0 \pmod16 $, so we only need the last 4 digits: $ 8024 $. Now $ 8024 \div 16 $: $ 16 imes 500 = 8000 $, $ 8024 - 8000 = 24 $, $ 24 \div 16 = 1 $ - Portal da Acústica