Question

$(49{)}^{15}-1$ is exactly divisible by:

• A. $50$
• B. $51$
• C. $29$
• D. $8$

Answer 1

The correct option is D $8$

$\to {49}^{15}-1=(1+48{)}^{15}-1$

$={\sum }_{r=0}^{15}$ ${}^{15}{C}_r(48{)}^r-1$

${=}^{15}{C}_0\left(48{)}^0{+}^{15}{C}_1\right(48{)}^1+...{+}^{15}{C}_{15}(48{)}^{15}-1$

${=}^{15}{C}_1(48{)}^1+--{+}^{15}{C}_{15}(48{)}^{15}$

$=\left(48\right){(}^{15}{C}_1+---{+}^{15}{C}_{15}\left(48{)}^1\right)$

$=48k\Rightarrow$ Divisible by 48

$\Rightarrow$ Divisible by 8 also

$\Rightarrow \left(D\right)$