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# Find the square of the following number without actual multiplication (i) 39 (ii)42

Last updated date: 20th Jun 2024
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Hint: Square- The positive number, when multiplied by itself, represents the square of the number.
Example : - a $\times$ a$=$${a^2}$
Where a $=$ any natural no.
In the example we can see that, when any number multiplied by itself then that is referred to as the square of that individual number.
Any natural number n which can be represented as ${n^2}$, where n is a natural number, then n is called a square number
In this question we can take the help of the formulas which are given below by solving the square without actual multiplication $\to$
$\begin{gathered} {(a - b)^2} = {a^2} + {b^2} - 2ab \\ {(a + b)^2} = {a^2} + {b^2} + 2ab \\ \end{gathered}$

Square of the number 39 without actual multiplication
$39 \times 39 = {(39)^2}$
We can also write as
${(39)^2} = {(40 - 1)^2}$
With the help of the formula
${(a - b)^2} = {a^2} + {b^2} - 2ab$
We should solve ${(40 - 1)^2}$
$\begin{gathered} a = 40,\,b = 1 \\ {(40 - 1)^2} = {(40)^2} + {(1)^2} - 2 \times 40 \times 1 \\ = 1600 + 1 - 80 \\ = 1601 - 80 \\ = 1521 \\ \end{gathered}$
ii) 42
$42 \times 42 = {(42)^2}$
We can also write as
${(42)^2} = {(40 + 2)^2}$
With the help of the formula
$\begin{gathered} {(a + b)^2} = {a^2} + {b^2} + 2ab \\ a = 40\,\,b = 2 \\ {(40 + 2)^2} = {(40)^2} + {(2)^2} + 2 \times 40 \times 2 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1600 + 4 + 160 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1760 + 4 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1764 \\ \end{gathered}$
Hence the square of 39 and 42 without actual multiplication is 1521 and 1764 respectively.

Note: Square numbers are the complete pair of its prime factor
There is short trick to find the square of a any number
${(39)^2}$
Base number $= 39$
First digit $= 3$
Last digit $= 9$
Step 1 $= 9 + 39 = 48$
Step 2 $= 48 \times 3 = 144$
Step 3 $= {(9)^2} = 81$
Step 4 $= 1440+81 = 1521$