Ufumbuzi - Tiger algebra Suluhishaji

1, 218

1,218

Maelezo kwa hatua

${(4 C 2)}_{16} = b a s e 10$

$4 C 2$ , $16$ , $10$ .

${(4 C 2)}_{16} = 4 \cdot 16^{2} + 12 \cdot 16^{1} + 2 \cdot 16^{0}$

$4 \cdot 16^{2} + 12 \cdot 16^{1} + 2 \cdot 16^{0} = 1218$

Hatua: $4 \cdot 16^{2}$
+ $12 \cdot 16^{1}$
+ $2 \cdot 16^{0}$ = $1, 218$ .

$\frac{1218}{10} = 121 r e m a \in d e r 8$

$\frac{121}{10} = 12 r e m a \in d e r 1$

$\frac{12}{10} = 1 r e m a \in d e r 2$

$\frac{1}{10} = 0 r e m a \in d e r 1$

$8 \to 1 \to 2 \to 1 = 1218$

$1218 (b a s e 10) = {(1218)}_{10}$

$1, 218_{10} = 1, 218_{10}$ .

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