Question

Consider $n={21}^{52}$, then

• A. number of even divisors of $n$ is $704$
• B. number of odd divisors of $n$ is $2809$
• C. last two digits of $n$ is $41$
• D. number of even divisors of $n$ which are multiple of $9$ is $2705$

Answer 1

Answer: (B), (C)

The correct options are
B number of odd divisors of $n$ is $2809$
C last two digits of $n$ is $41$

Let,we have

$n={21}^{52}$

can be written as $n={\left(7\times 3\right)}^{52}$

$n=7^{52}{.3}^{52}$

We know, no. of total divisors of any number

$k=p^m.q^n$

Total divisors$=(m+1)(n+1)$

so, similary here

odd divisors$=(52+1)(52+1)=2809$

Hence the option $\left(B\right)$ is correct

But again

For last two digit

$n={21}^{52}={\left(20+1\right)}^{52}$

${\left(20+1\right)}^{52}{=}^{52}{C}_1{\left(20\right)}^{52}+.....{+}^{52}{C}_{51}{\left(20\right)}^1{+}^{52}{C}_{52}{\left(20\right)}^0$

For last two digit we notice last two terms

${=}^{52}{C}_{51}\left(20\right)+1$

$=52\times 20\times 1$

$=1041$

$1041$ has last two digit is $41$

so, option $\left(C\right)$ is also correct

Hence both the option $\left(B\right)$ and $\left(C\right)$ are correct.