Question

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

Answer 1

Answer: 3

${\left(1+5x+10x^2+10x^3+5x^4+x^5\right)}^{20}$
$={\left((1+x{)}^5\right)}^{20}=(1+x{)}^{100}$
Number of terms $=101=\lambda$ $3^{101}=3\cdot 3^{100}=3\cdot 9^{50}=3(10-1{)}^{50}$
$=3({}^{50}{C}_0\cdot 10^{50}-{}^{50}{C}_1\cdot 10^{49}+\dots \dots \dots +{}^{50}{C}_{48}\cdot 10^2-{}^{50}{C}_{49}\cdot 10+{}^{50}{C}_{50})$
$=3\left(100K+\frac{50\times 49}{2}\times 100-50\times 10+1\right)$
$=3K\times 100+3\times 25\times 49\times 100-500\times 3+3$
$=$ Multiple of $100+3$
$\Rightarrow$ one's place is 3 and ten's place is 0 Hence, $O+T=3$