Number Operations and Properties

$29\times31$ $=$ $(30-1)$ $(30+1)$

$16\times18$ $=$ $(20-4)$ $(20-2)$

$22\times 24=(20+2)$$\times$$(20+4)$ 

$19\times 21=(20-1)$ $\times$ $(20+1)$ 

$14\times 16=(10+4)$$\times$$(10+6)$ 

$185$ $=$ $(200-15)$

$31\times 33=(30+1)$$\times$$(30+3)$ 

$11^{2}$ $=$ $(10+1)^{2}$ $=$ $100+1+20$

$21^{2}-1 = 400$
we can express the above number into general form $a^{2}-1 = (a + 1)\times (a - 1)$
Where a = 21
So, $21^{2}-1 = (21 + 1)\times (21 - 1)$
= $22 \times 20 = 400$
Therefore, the two even consecutive numbers are 20 and 22.

$11^{2}-1 = 120$
we can express the above number into general form $a^{2}-1 = (a + 1)\times (a - 1)$
Where a = 11
So, $11^{2}-1 = (11 + 1)\times (11 - 1)$
= $12 \times 10 = 120$
Therefore, the two even consecutive numbers are 10 and 12.

From the pattern, the third number is the difference of the first two numbers.
The fourth number can be obtained by addition of the first two numbers.
Then, the missing numbers will be
$14^2 + 7^2 + 7^2 = 21^2$
So, $7^2, 21^2$ are the missing numbers.

when we divide $100$ with $99$ we will get remainder $1$ and hence next we divide 99 with $1$ which gives us remainder $0$.

$-9 - (-5)$

$12r-11r$

$19y-15y$