Question

A gardener has \[1400\] plants. He wants to plant these in such a way that the number of rows and the number of columns remains the same. Find the minimum number of plants he needs more for this.

Answer 1

Hint: First we have to know what is a square root of a number. We used the method of finding the square root of any number and finding the minimum number of plants that add to the given number of plants to plant those in such a way that the number of rows and the number of columns remains the same.

Complete step by step solution:
Let \[x\] be the number of rows.
Given a garden has \[1400\] plants. Suppose the number of rows and number of columns remains the same. Then the number of columns is \[x\].
Hence the total number of plants required is \[{x^2}\]. In the given number, mark off the digits in pairs starting from the unit’s digits. Each pair and remaining one digit is called a period.
Now, \[{3^2} = 9\]. On subtracting, we get \[5\]as remainder. then bring the down next period i.e., \[00\] . Now, trial divisor is \[3 + 3 = 6\] and trial dividend is \[500\]. We want to add the number to get the perfect square. So, we take \[38\] as a divisor and put \[8\]as quotient. The remainder is \[ - 44\]. Hence, we have to add \[44\] to get the perfect square. i.e.,
         \[\begin{gathered}
  \;\,\begin{array}{*{20}{c}}
  3 \\
  {}
\end{array}\left| \!{\underline {\,
  \begin{gathered}
  14\;00\;(38 \\
 9 \,\,\,\, \\
\end{gathered} \,}} \right. \\
  \begin{array}{*{20}{c}}
  {68} \\
  {}
\end{array}\left| \!{\underline {\,
  \begin{gathered}
  500 \\
  544 \\
\end{gathered} \,}} \right. \\
  \quad \; - 44 \\
\end{gathered} \]
Hence, the minimum number of plants is \[44\].

Note:
Note that if \[{x^2} = y\], we say that the square root of \[y\] is \[x\] and we write, \[\sqrt y = x\]. The cube root of a given number \[x\] is the number whose cube is \[x\]. We denote the cube root of \[x\] by \[\sqrt[3]{x}\].