Thinking Mathematically (6th Edition)

$44,261$
RECALL: in the hexadecimal system. $A= 10$, $C=12$, and $E=14$. Thus, the expanded form of $ACE5_{\text{sixteen}}$ is: $=(10 \times 16^3) + (12 \times 16^2)+(14 \times 16^1) + (5 \times 1) \\=(10 \times 4096) + (12 \times 256) + (14 \times 16) + 5 \\=40,960+3,072 + 224+5 \\=44,261$