When I was a young boy I was shown the following trick to square certain numbers very quickly. In particular, given a 2-digit number where the unit digit is 5, the square of this number can be easily computed with the following rule:
If the tens digit is a and the unit digit is 5, the square can be found by computing a(a+1) and appending 25 to the result.
For instance 752 is computed by multiplying 7 and 8 (=56)and adding 25 at the end to get 5625. This also work with 3-digit numbers (and longer numbers). For instance, to compute 1152, we compute 11×12=132 and get 13225.
To show why this is true is a simple exercise. A number with tens digit a and unit digit 5 is equal to 10a+5. Squaring this number results in (10a+5)2=100a2+100a+25=100a(a+1)+25 which corresponds to the rule above.
This same argument is true when multiplying two 2-digit numbers where the tens digits are equal and the unit digits add up to 10. In this case we are multiplying 10a+b and 10a+c where b+c=10. The product (10a+b)(10a+c) is equal to 100a2+100a+bc=100a(a+1)+bc. So the corresponding rule is:
Multiply a with a+1 and append bc at the end.
The only caveat is that when b=1 and c=9, you append 09 (rather than just 9) at the end. For instance to compute 83×87, we get 8×9=72 and append 3×7 to get 7221. Similarly for 124×126, we compute 12×13=156 and append 24 to get 15624.
You can use a similar trick to multiply two 2-digit number where the tens digits add up to 10 and the unit digits are the same, i.e. multiplying 10b+a with 10c+a where b+c=10. The product is
(10b+a)(10c+a)=100bc+100a+a2=100(bc+a)+a2
Thus the corresponding rule is:
Multiply b with c, add a and then append a2 to the result.
Again, if a2 has only one digit, prefix it with a 0 before appending it to bc+a. For example, to compute 43×63, we compute 4×6+3=27 and append 32=09 to obtain 2709.
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