Question

A four digit perfect square number whose first two digits and last two digits taken seperately are also perfect squares is:

• A. 1,681
• B. 1,636
• C. 3,664
• D. 6,481

Answer 1

Answer: (A)

1,681

$1,681$ is the only perfect square from given options.

$1681=41\times 41$

$1681={41}^2$

$\therefore$ $1681$ is a square of $41$.

First two digits and last two digits of $1681$ are $16$ and $81$

$16=4^2$ and $81=9^2$

We can see $16$ and $81$ are perfect squares of $4$ and $9$.

$\therefore$ $1681$ is a four digit square number whose first two digits taken separately are also perfect square.