Question

Find the value of $\left\{\frac{3^{2003}}{28}\right\}$, where {$\cdot$} denotes the fractional part.

• A. $\frac{9}{28}$
• B. $\frac{10}{28}$
• C. $\frac{19}{28}$
• D. $\frac{20}{28}$

Answer 1

The correct option is C $\frac{19}{28}$

$E=3^{2003}=3^{2001}\times 3^2=9(27{)}^{667}=9(28-1{)}^{667}$
$⟹E=9{[}^{667}{C}^o{28}^{667}{-}^{667}{C}^1(28{)}^{666}+...{-}^{667}{C}_{667}]=9\times 28k-9or\frac{E}{28}=9k-\frac{9}{28}=9k-1+\frac{19}{28}$
That means if we dived 3${}^{2003}$ by $28$, the remainder is $19$. Thus,
$\left\{\frac{3^{2003}}{28}\right\}=\frac{19}{28}$