Hi,
I there any easy way to find suare of numbers between 40 and 50, 60 and 90?
Thanks!
Well $\displaystyle 6\times6=36$ and $\displaystyle 10\times10=100$, that leaves you to try 7,8 and 9.
Algebraically?
I suppose you could say that
$\displaystyle 40<x^2<50$
$\displaystyle \sqrt{40}<x<\sqrt50$
$\displaystyle 6.32<x<7.07$
So 7 squared lies between 40 and 50, and adapt this for the other numbers.
Hello, ManuLi!
Is there any easy way to find squares of numbers between 40 and 50, 60 and 90?
There is a trick for squaring numbers "near 50",
Let $\displaystyle N \:=\:50 \pm d$ .where $\displaystyle d$ is the "deviation from 50".
Then: .$\displaystyle N^2 \:=\:(50 \pm d)^2 \:=\:2500 \pm 100d + d\:\!^2 \:=\:100(25 \pm d) + d\:\!^2 $
So, in the hundreds-place is: .$\displaystyle 25 \pm d$
. . . followed by $\displaystyle d\:\!^2$, the deviation-squared.
Example: .$\displaystyle 53^2$
The deviation is: .$\displaystyle d = +3$
The answer is: .$\displaystyle 25 + 3 \:=\:28$
. .followed by: .$\displaystyle 3^2 \:=\:09$
Therefore: .$\displaystyle 53^2 \:=\:2809$
Example: .$\displaystyle 44^2$
The deviation is: .$\displaystyle d = -6$
The answer is: .$\displaystyle 25 - 6 \:=\:19$
. .followed by: .$\displaystyle (\text{-}6)^2 \:=\:36$
Therefore: .$\displaystyle 44^2 \;=\;1936$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Consecutive squares differ by consecutive odd numbers.
. . $\displaystyle \begin{array}{cccccccccccccc}
25 && 36 && 49 && 64 && 81 && 100 && \hdots\\
& 11 && 13 && 15 && 17 && 19 && \hdots \end{array}$
If we know one square, we can crank out the next few squares.
For example: .$\displaystyle 60^2 \,=\,3600$
Double the 60 and add 1: .$\displaystyle 2(60) + 1 \,=\,121 $
That is the odd number we will add.
. . $\displaystyle \begin{array}{ccc}\;3600 &=& 60^2 \\
+ 121 \\
\;3721 &=& 61^2 \\
+ 123 \\
\;3844 &=& 62^2 \\
+ 125 \\
\;3969 &=& 63^2 \\
+ 127 \\
\;4096 &=& 64^2 \\
\vdots && \vdots
\end{array}$