Question

Let $\lambda $ denote the number of terms in the expansion of $\left(1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5}\right)^{20}$ . If unit’s place and ten’s place digits in $3^{\lambda }$ are $O$ and $T$ , then $O+T$ is equal to

Answer 1

$\left(1+5 x+10 x^{2}+10 x^{3}+5 x^{4}+x^{5}\right)^{20}$
$=\left((1+x)^{5}\right)^{20}=(1+x)^{100}$
Number of terms $=101=\lambda$ $3^{101}=3 \cdot 3^{100}=3 \cdot 9^{50}=3(10-1)^{50}$
$=3\left({ }^{50} C_{0} \cdot 10^{50}-{ }^{50} C_{1} \cdot 10^{49}+\ldots \ldots \ldots+{ }^{50} C_{48} \cdot 10^{2}-{ }^{50} C_{49} \cdot 10+{ }^{50} C_{50}\right.)$
$=3\left(100 K+\frac{50 \times 49}{2} \times 100-50 \times 10+1\right)$
$=3 K \times 100+3 \times 25 \times 49 \times 100-500 \times 3+3$
$=$ Multiple of $100+3$
$\Rightarrow $ one's place is 3 and ten's place is 0 Hence, $O+T=3$