Question

The value of $\left\{\frac{3^{2003}}{28}\right\}$ is
(where {.} denotes the fractional part function)

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

Answer 1

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

$3^{2003}=3^2\times 3^{2001}=9(3^3{)}^{667}=9(27{)}^{667}=9(28-1{)}^{667}=9[{}^{667}{C}_0{28}^{667}-{}^{667}{C}_1(28{)}^{666}+\dots {-}^{667}{C}_{667}]=9[28k-1]=9\times 28k-9$

Now,
$\frac{3^{2003}}{28}=9k-\frac{9}{28}=9k-1+\frac{19}{28}$
As $k\in N$, so $9k-1\in N$

Hence, $\left\{\frac{3^{2003}}{28}\right\}=\frac{19}{28}$