# Squaring numbers

• Dec 29th 2010, 04:26 PM
Hellbent
Squaring numbers
Hi,

How is the part after the arrow arrived at?

$\displaystyle \begin{array}{rlc}84 \cdot 90 \rightarrow 10(80 \cdot 9 + 4 \cdot 9)\end{array}$
• Dec 29th 2010, 04:39 PM
Quote:

Originally Posted by Hellbent
Hi,

How is the part after the arrow arrived at?

$\displaystyle \begin{array}{rlc}84 \cdot 90 \rightarrow 10(80 \cdot 9 + 4 \cdot 9)\end{array}$

$\displaystyle 84*90=10[84*9]=10[(80+4)*9]=10[80*9+4*9]$
• Dec 29th 2010, 05:11 PM
Hellbent
$\displaystyle \mathrm{What\: if\ I\ wanted\ to\ try\ 83 \cdot 91?}$
Factoring doesn't work, does it?
• Dec 29th 2010, 05:14 PM
Quote:

Originally Posted by Hellbent
$\displaystyle \mathrm{What\: if\ I\ wanted\ to\ try\ 83 \cdot 91?}$
Factoring doesn't work, does it?

You could write it as

$\displaystyle (83)(91)=(80+3)(90+1)=(80)(90)+(80)(1)+(3)(90)+(3) (1)$
• Dec 29th 2010, 05:28 PM
Soroban
Hello, Hellbent!

Quote:

$\displaystyle \text{What if I wanted to try } 83\cdot91\,?$
$\displaystyle \text{Factoring doesn't work, does it?}$

Here's one way . . .

$\displaystyle \begin{array}{ccccccc}83\cdot91 & \Rightarrow & 83\cdot(7\!\cdot\!13) \\ \\[-3mm] & \Rightarrow & 7\cdot(83\!\cdot\!13) \\ \\[-3mm] & \Rightarrow & 7\cdot\bigg[(80+3)\!\cdot\!13\bigg] \\ \\[-3mm] & \Rightarrow & 7\cdot\bigg[80\!\cdot\!13 + 3\!\cdot\!13\bigg] \end{array}$

• Dec 29th 2010, 05:36 PM
$\displaystyle \mathrm{What\: if\ I\ wanted\ to\ try\ 83 \cdot 91?}$
You can also use your result from $\displaystyle (84)(90)$
$\displaystyle (83)(91)=(84-1)(90+1)=(84)(90)+84-90$
$\displaystyle =10[80*9+4*9]-6$