Question

Find the least number that must be subtracted from 5607 so as to get a perfect square. Also, find the square root of the perfect square.

Answer 1

Hint:
Here, we will try to find the square root of 5607 by division method. If we get the remainder as 0, then 5607 will be a perfect square. However, if the remainder is any number other than 0, then it will not be a perfect square. Subtracting the remainder from 5607 will give us the required number which is a perfect square. The remainder obtained will be the least number which must be subtracted from 5607 to make it a perfect square.

Complete step by step solution:
First, we will try to find the square root of the given number 5607 by using Division method.

\[\begin{array}{l}\begin{array}{*{20}{c}}{}&{74}\end{array}\\7\left| \!{\overline {\,
 \begin{array}{l}\overline {56} \overline {07} \\ - 49\end{array} \,}} \right. \\\begin{array}{*{20}{l}}{144}\\{}\\{}\end{array}\left| \!{\overline {\,
 \begin{array}{l}707\\\underline { - 576} \\131\end{array} \,}} \right. \end{array}\]
Therefore, 131 is the required least number.
Now, we will subtract the remainder i.e. 131 from the given number.
Hence, we get,
\[\left( {5607 - 131} \right) = 5476\]
Now, we will find the square root of this number either by the division method above or by doing its prime factorization.
Prime factorization of 5476 is:
$\begin{array}{*{20}{l}}
     2 | & {5476} \\
\hline
     2 | & {2738} \\
\hline
{37} | & {1369} \\
\hline
{37} | & {37} \\
\hline
    {} | & 1
\end{array}$
Hence, 5476 can be written as:
\[5476 = 2 \times 2 \times 37 \times 37\]
Now, we are required to find the square root, so we will take only one prime number out of a pair of the same prime numbers.
\[ \Rightarrow \sqrt {5476} = 2 \times 37\]
Multiplying the terms, we get
\[ \Rightarrow \sqrt {5476} = 74\]

Hence, the square root of 5476 is 74.

Note:
In the division method, we start pairing the numbers by taking a bar starting from the unit’s place. We take the largest possible number whose square will be less than or equal to the leftmost pair.
Now, we double the value of the quotient and assume it as a ten’s place digit. For the one place we have to take a digit such that multiplying the total number by that digit, we get a number less than or equal to the carried down number. Since, in this case, the remainder is 131 and not 0 hence, 131 is the least number which should be subtracted from 5607 to make it a perfect square and hence, get the remainder 0.