Question

For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) $252$ (ii) $2925$ (iii) $396$ (iv) $2645$ (v) $2800$ (vi) $1620$

Answer 1

(i) $252$ can be factorized as follows,

$252=\underset{–}{2\times 2}\times \underset{–}{3\times 3}\times 7$

Here, prime factor $7$ does not have its pair.

If we divide this number by $7$, then the number will become a perfect square. Therefore, $252$ has to be divided by $7$ to obtain a perfect square.

$\frac{252}{7}=36$ is a perfect square as,

$36=\underset{–}{2\times 2}\times \underset{–}{3\times 3}=6\times 6=6^2$

(ii)

$2925$ can be factorized as follows.

$2925=\underset{–}{3\times 3}\times \underset{–}{5\times 5}\times 13$

Here, prime factor $13$ does not have its pair.

If we divide this number by $13$, then the number will become a perfect square. Therefore, $2925$ has to be divided by $13$ to obtain a perfect square.

$\frac{2925}{13}=225$ is a perfect square as,

$225=\underset{–}{3\times 3}\times \underset{–}{5\times 5}=15\times 15$

(iii)

$396$ can be factorized as follows.

$396=\underset{–}{2\times 2}\times \underset{–}{3\times 3}\times 11$

Here, prime factor $11$ does not have its pair.

If we divide this number by $11$, then the number will become a perfect square. Therefore, $396$ has to be divided by $11$ to obtain a perfect square.

$\frac{396}{11}=36$ is a perfect square as,

$36=\underset{–}{2\times 2}\times \underset{–}{3\times 3}=6\times 6=6^2$

(iv)

$2645$ can be factorized as follows.

$2645=\underset{–}{23\times 23}\times 5$

Here, prime factor $5$ does not have its pair.

If we divide this number by $5$, then the number will become a perfect square.

Therefore, $2645$ has to be divided by $5$ to obtain a perfect square.

$\frac{2645}{5}=529$ is a perfect square.

$529=\underset{–}{23\times 23}=13\times 13$

(v)

$2800$ can be factorized as follows.

$2800=\underset{–}{2\times 2}\times \underset{–}{2\times 2}\times \underset{–}{5\times 5}\times 7$

Here, prime factor $7$ does not have its pair.

If we divide this number by $7$, then the number will become a perfect square.

Therefore, $2800$ has to be divided by $7$ to obtain a perfect square.

$\frac{2800}{7}=400$ is a perfect square.

$400=\underset{–}{2\times 2}\times \underset{–}{2\times 2}\times \underset{–}{5\times 5}=20\times 20$

(vi)

$1620$ can be factorized as follows.

$1620=\underset{–}{2\times 2}\times \underset{–}{3\times 3}\times \underset{–}{3\times 3}\times 5$

Here, prime factor $5$ does not have its pair.

If we divide this number by $5$, then the number will become a perfect square.

Therefore,$1620$ has to be divided by $5$ to obtain a perfect square.

$\frac{1620}{5}=324$is a perfect square.

$\Rightarrow 324=\underset{–}{2\times 2}\times \underset{–}{3\times 3}\times \underset{–}{3\times 3}=18\times 18$