Solution - Base conversion

2, 498

2,498

Step-by-step explanation

${(9 C 2)}_{16} = b a s e 10$

Identify the source number, source base, and target base before converting through base 10.

${(9 C 2)}_{16} = 9 \cdot 16^{2} + 12 \cdot 16^{1} + 2 \cdot 16^{0}$

$9 \cdot 16^{2} + 12 \cdot 16^{1} + 2 \cdot 16^{0} = 2498$

Expand each digit by place value: $9 \cdot 16^{2}$
+ $12 \cdot 16^{1}$
+ $2 \cdot 16^{0}$ . This gives $2, 498$ in base 10.

$\frac{2498}{10} = 249 r e m a \in d e r 8$

$\frac{249}{10} = 24 r e m a \in d e r 9$

$\frac{24}{10} = 2 r e m a \in d e r 4$

$\frac{2}{10} = 0 r e m a \in d e r 2$

$8 \to 9 \to 4 \to 2 = 2498$

$2498 (b a s e 10) = {(2498)}_{10}$

Repeatedly divide $2, 498$ by $10$ , collect remainders, and read them in reverse to get $2, 498$ .

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Base conversion is foundational for computer science, digital systems, and number theory.