In the posting this time explains how fast looking for find quadratic results number two digits.
Earlier we've definitely know how to conventional search for a result quadratic of a number two digits.
But different in a manner that the following. Here is a formula essentially:
$(ab)^2=a^2 \quad a(b)(2) \quad b^2$
Where $a$ and $b$ are the original number.
Now Look at this example well!
Example 1:
$13^2=1^2 \quad 1(3)(2) \quad 3^2$ $=169$
Example 2:
$27^2=2^2 \quad 2(7)(2) \quad 7^2$ $=4 \quad 28 \quad 49$
Remember the summation technique with the way down, than:
$= 4 \quad (28+4=32) \quad 9$
$=4+3 \quad 2 \quad 9$
So, $27^2=729$
Example 3:
$87^2=64,112,49$ $=64,116,9=7569$
So, $87^2=7569$
Example 4:
$79^2=49,126,81$ $=49,134,1=6241$
So, $79^2=6241$
Ok friends, thank you for your hobby you guys with math, to see you and good luck.
$(ab)^2=a^2 \quad a(b)(2) \quad b^2$
Where $a$ and $b$ are the original number.
Now Look at this example well!
Example 1:
$13^2=1^2 \quad 1(3)(2) \quad 3^2$ $=169$
Example 2:
$27^2=2^2 \quad 2(7)(2) \quad 7^2$ $=4 \quad 28 \quad 49$
Remember the summation technique with the way down, than:
$= 4 \quad (28+4=32) \quad 9$
$=4+3 \quad 2 \quad 9$
So, $27^2=729$
Example 3:
$87^2=64,112,49$ $=64,116,9=7569$
So, $87^2=7569$
Example 4:
$79^2=49,126,81$ $=49,134,1=6241$
So, $79^2=6241$
Ok friends, thank you for your hobby you guys with math, to see you and good luck.