1. ## Finding Square

Hi,

I there any easy way to find suare of numbers between 40 and 50, 60 and 90?

Thanks!

2. Well $6\times6=36$ and $10\times10=100$, that leaves you to try 7,8 and 9.

Algebraically?

I suppose you could say that
$40
$\sqrt{40}
$6.32
So 7 squared lies between 40 and 50, and adapt this for the other numbers.

3. He was asking for squares, not square roots.

ManuLi, you can do things like 47= 40+ 7 so $47^2= 40^2+ 2(40)(7)+ 7^2= 1600+ 560+ 49$ but I don't know that that is any simpler than just multiplying 47 times 47.

4. Oh I misunderstood, evidently. Sorry for the useless post .

5. I'd try a spreadsheet. Easy to set up a sequence, and they are all there instantly.

6. 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 $N \:=\:50 \pm d$ .where $d$ is the "deviation from 50".

Then: . $N^2 \:=\:(50 \pm d)^2 \:=\:2500 \pm 100d + d\:\!^2 \:=\:100(25 \pm d) + d\:\!^2$

So, in the hundreds-place is: . $25 \pm d$
. . . followed by $d\:\!^2$, the deviation-squared.

Example: . $53^2$

The deviation is: . $d = +3$

The answer is: . $25 + 3 \:=\:28$
. .followed by: . $3^2 \:=\:09$

Therefore: . $53^2 \:=\:2809$

Example: . $44^2$

The deviation is: . $d = -6$

The answer is: . $25 - 6 \:=\:19$
. .followed by: . $(\text{-}6)^2 \:=\:36$

Therefore: . $44^2 \;=\;1936$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Consecutive squares differ by consecutive odd numbers.

. . $\begin{array}{cccccccccccccc}
25 && 36 && 49 && 64 && 81 && 100 && \hdots\\
& 11 && 13 && 15 && 17 && 19 && \hdots \end{array}$

If we know one square, we can crank out the next few squares.

For example: . $60^2 \,=\,3600$

Double the 60 and add 1: . $2(60) + 1 \,=\,121$
That is the odd number we will add.

. . $\begin{array}{ccc}\;3600 &=& 60^2 \\
+ 121 \\
\;3721 &=& 61^2 \\
+ 123 \\
\;3844 &=& 62^2 \\
+ 125 \\
\;3969 &=& 63^2 \\
+ 127 \\
\;4096 &=& 64^2 \\
\vdots && \vdots
\end{array}$