The number of integers $q , 1 \leq q \leq 2021$, such that $\sqrt{ q }$ is rational, and $\frac{1}{ q }$ has a terminating decimal expansion, is
The number of integers $q , 1 \leq q \leq 2021$, such that $\sqrt{ q }$ is rational, and $\frac{1}{ q }$ has a terminating decimal expansion, is
- [KVPY 2021]
- A
$1$
- B
$11$
- C
$22$
- D
$44$
Similar Questions
If ${({a^m})^n} = {a^{{m^n}}}$, then the value of $'m'$ in terms of $'n'$ is
If ${({a^m})^n} = {a^{{m^n}}}$, then the value of $'m'$ in terms of $'n'$ is
If ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ then $m =$
If ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ then $m =$
Let ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}}$$ = A + B{.2^{1/4}} + C{.2^{1/2}} + D{.2^{3/4}}$, then $A+B+C+D= . . .$
Let ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}}$$ = A + B{.2^{1/4}} + C{.2^{1/2}} + D{.2^{3/4}}$, then $A+B+C+D= . . .$
${a^{m{{\log }_a}n}} = $
${a^{m{{\log }_a}n}} = $
$\sqrt {(3 + \sqrt 5 )} - \sqrt {(2 + \sqrt 3 )} = $
$\sqrt {(3 + \sqrt 5 )} - \sqrt {(2 + \sqrt 3 )} = $