Question

Find whether the square of the following numbers are even or odd?
$9999$

Answer 1

here $9999=10000-1$

now square of $a-b$ can be written as $a^2-2ab+b^2$

So, $(9999{)}^2=(10000-1{)}^2$

$=(10000{)}^2-2(10000\left)\right(1)+(1{)}^2$

$\to$ now, any even number square became even number only

$\to$ Multiplication of any number with even number converts into even it self

$\to$ here $(1{)}^2$ will be odd number.

So, $(9999{)}^2=EVEN+(-EVEN)+ODD$

$\to$ We can say that $EVEN+EVEN=EVEN$

$EVEN+ODD=ODD$

$ODD+ODD=EVEN$

So, $(9999{)}^2=EVEN-EVEN+ODD$

$=EVEN+ODD$

$(9999{)}^2=ODD$