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.

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