20-21 question

Let $\begin{aligned} \xi &= 2020^{2021n} + 2021^{2020n} \\ \implies \xi &= 2020^{2021n} \Big( 1 + \frac{2021^{2020n} }{ 2020^{2021n} } \Big) \\ \implies \xi &= 2020^{2021n} \Big[ 1 + \Big( \frac{2021^{2020} }{ 2020^{2021} } \Big)^n \Big] \\ \implies \xi &= 2020^{2021n} (1 + \phi^n) \end{aligned}$

where $\phi = \frac{2021^{2020} }{ 2020^{2021} } < 1$

So, $\begin{aligned} \Lambda = \lim_{n \to \infty} \xi^{\frac{1}{n}} &= \lim_{n \to \infty} [2020^{2021n} (1 + \phi^n)]^{\frac{1}{n}} \\ &= 2020^{2021} [ \lim_{n \to \infty} (1 + \phi^n)^{\frac{1}{n}} ] \\ &= 2020^{2021} [ \lim_{n \to \infty} (1 + \phi^n) ]^{\frac{1}{n}} \\ &= 2020^{2021} ( 1 + \lim_{n \to \infty} \phi^n )^{\frac{1}{n}} \\ &= 2020^{2021} ( 1 + 0 )^{\frac{1}{n}} \\ \therefore \Lambda &= 2020^{2021} \end{aligned}$

If a decimal number has the form $10^k$ , it has $\lfloor k \rfloor + 1$ digits. $\Lambda = 2020^{2021} = 10^{ 2021 \log 2020 } \approx 10^{6680.115}$

Therefore, $\Lambda$ has $\boxed{6681}$ digits.

N.B. Why $\phi < 1$ $\begin{aligned} 2020^{2021} \> &? \> 2021^{2020} \\ 2020 \cdot 2020^{2020} \> &? \> 2021^{2020} \\ 2020 \> &? \> \Big(\frac{2021}{2020}\Big)^{2020} \\ 2020 \> &? \> \Big(1+ \frac{1}{2020}\Big)^{2020} \end{aligned}$

Now, $e = \lim_{x \to \infty} \Big(1+\frac{1}{x}\Big)^x > \Big(1+ \frac{1}{2020}\Big)^{2020}$

So, $2020 > e > \Big(1+ \frac{1}{2020}\Big)^{2020}$ $\implies 2020^{2021} > 2021^{2020}$

$\begin{aligned} \Lambda & = \lim_{n \to \infty} \sqrt[n]{2020^{2021n} + 2021^{2020n}} \\ & = \lim_{n \to \infty} \exp \left(\frac 1n \ln \left(2020^{2021n} + 2021^{2020n} \right) \right) \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln \left(2020^{2021n} + 2021^{2020n} \right)}n \right) \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln \left(2020^{2021n} \left(1 + 2020^{-2021n}2021^{2020n} \right)\right)}n \right) \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln \left(2020^{2021n} \right) + \ln \left(1 + 2020^{-2021n}2021^{2020n} \right)}n \right) \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln \left(2020^{2021n} \right)}n + \lim_{n \to \infty} \frac {\ln \left(1+ \frac 1{2020^n} \cdot \left(1 + \frac 1{2020}\right)^{2020n} \right)}n \right) \\ & = \exp \left(\lim_{n \to \infty} \frac {2021n \ln 2020}n + \lim_{n \to \infty} \frac {\ln \left(1+ \left(\frac {(1+1/2020)^{2020}}{2020} \right)^n \right)}n \right) \\ & = \exp \left(2021 \ln 2020 + \frac {\ln (1+0)}\infty \right) \\ & = 2020^{2021} \end{aligned}$

Number of digits of $\Lambda$ is given by $\left \lfloor \log_{10} \Lambda \right \rfloor + 1 = \left \lfloor 2021 \log_{10} 2020 \right \rfloor + 1 = 6680 + 1 = \boxed{6681}$ .

We can rewrite the inside part of the root:

$2020^{2021n}+2021^{2020n}=2020^{(2020+1)n}+(2020+1)^{2020n}$

For simplicity and easier reading, let $a=2020$ . By the binomial theorem:

$(a+1)^{an}=\sum_{i=0}^{an} {an\choose i} a^{an-i}$ $\implies a^{(a+1)n}+(a+1)^{an}=a^{(a+1)n}+\sum_{i=0}^{an}{an\choose i}a^{an-i}$ $=a^{(a+1)n}\left[1+\sum_{i=0}^{an}{an\choose i}a^{-(n+i)}\right]$

We plug this into the root:

$\Lambda=\lim_{n\rightarrow\infty}\sqrt[n]{a^{(a+1)n}\left(1+\sum_{i=0}^{an}{an\choose i}a^{-(n+i)}\right)}$ $=\lim_{n\rightarrow\infty}a^{(a+1)}\sqrt[n]{1+\sum_{i=0}^{an}{an\choose i}a^{-(n+i)}}$

We take the log (base 10):

$\log\Lambda=\log\left(\lim_{n\rightarrow\infty}a^{(a+1)}\sqrt[n]{1+\sum_{i=0}^{an}{an\choose i}a^{-(n+i)}}\right)$ $=\lim_{n\rightarrow\infty}\log\left(a^{(a+1)}\sqrt[n]{1+\sum_{i=0}^{an}{an\choose i}a^{-(n+i)}}\right)$ $=(a+1)\log{a}+\lim_{n\rightarrow\infty}\frac{1}{n}\log\left({1+\sum_{i=0}^{an}{an\choose i}a^{-(n+i)}}\right)$ $=(a+1)\log{a}+\lim_{n\rightarrow\infty}\frac{1}{n}\log\left({1+\frac{1}{a^n}\sum_{i=0}^{an}{an\choose i}a^{-i}}\right)$ $=(a+1)\log{a}+\lim_{n\rightarrow\infty}\frac{1}{n}\log\left({1+\frac{1}{a^n}\left(1+\frac{1}{a}\right)^{an}}\right)$ $=(a+1)\log{a}+\lim_{n\rightarrow\infty}\frac{1}{n}\log\left({1+\left(\frac{\left(1+\frac{1}{a}\right)^a}{a}\right)^n}\right)$ $=(a+1)\log a$ $=2021\log2020$ $\implies\Lambda=2020^{2021}$

The number of digits is given by $\lfloor\log\Lambda\rfloor+1=\boxed{6681}$