1. Multiplication by 5
   It's often more convenient instead of multiplying by 5 to multiply first by 10 and then divide by 2.
   For example,

   137·5 = 1370/2 = 685.

   
   

2. Division by 5
   Similarly, it's often more convenient instead to multiply first by 2 and then divide by 10.
   For example,

   1375/5 = 2750/10 = 275.

   
   

3. Division/multiplication by 4
   Replace either with a repeated operation by 2.
   For example,

   124/4 = 62/2 = 31. Also, 124·4 = 248·2 = 496.

   
   

4. Division/multiplication by 25
   Use operations with 4 instead.
   For example,

   37·25 = 3700/4 = 1850/2 = 925.

   
   

5. Division/multiplication by 8
   Replace either with a repeated operation by 2.
   For example,

   124·8 = 248·4 = 496·2 = 992.

   
   

6. Division/multiplication by 125
   Use operations with 8 instead.
   For example,

   37·125 = 37000/8 = 18500/4 = 9250/2 = 4625.

   
   

7. Squaring two digit numbers.
   

   1. You should memorize the first 25 squares:

      1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 4 9 16 25 36 49 64 81 100 121 144 169 196 15 16 17 18 19 20 21 22 23 24 25 225 256 289 324 361 400 441 484 529 576 625

   2. Squares of numbers from 26 through 50.
      Let A be such a number. Subtract 25 from A to get x. Subtract x from 25 to get, say, a. Then A² = a² + 100x. For example, if A = 26, then x = 1 and 1375/5 = 2750/10 = 275. Hence

      262 = 242 + 100 = 676.

      
      

      Similarly, if A = 37, then x = 37 - 25 = 12, and a = 25 - 12 = 13. Therefore,

      372 = 132 + 100·12 = 1200 + 169 = 1369.

      Why does this work?

      (25 + x)2 - (25 - x)2 = [(25 + x) + (25 - x)]·[(25 + x) - (25 - x)]   = 50·2x   = 100x.

      
      

   3. Squares of numbers from 51 through 99.
      The idea is the same as above.

      (50 + x)2 - (50 - x)2 = 100·2x   = 200x.

      For example,

      32 = 372 + 200·13 = 1369 + 2600 = 3969.

      
      

   4. Squares of numbers from 51 through 99, second approach (this one was communicated to me by my late father Moisey Bogomolny).
      We are looking to compute A², where A = 50 + a. Instead compute 100·(25 + a) and add a². Example: 57². a = 57 - 50 = 7. 25 + 7 = 32. Append 49 = 7². Answer: 57² = 3249.

   5. Squares can be computed sequentially: (a + 1)² = a² + a + (a + 1). For example,

      1112 = 1102 + 110 + 111   = 12100 + 221   = 12321.

      
      

   6. In general, a² = (a + b)(a - b) + b². Let a be 57 and, again, we wish to compute 57². Let b = 3. Then

      572 = (57 + 3)(57 - 3) + 32,

      or

      572 = 60·54 + 9   = 3240 + 9   = 3249.

   
   

8. Squares of numbers that end with 5.
   Let A = 10a + 5. Then

   A2 = (10a + 5)2 = 100a2 + 2·10a·5 + 25 = 100a(a + 1) + 25.

   For example, compute 115², where a = 11. First compute 11·(11 + 1) = 11·12 = 132 (since 3 = 1 + 2). Next, append 25 to the right of 132 to get 13225! Another example: to compute 245² let a = 24. Then

   24·(24 + 1) = 242 + 24 = 576 + 24 = 600.

   Therefore 245²=60025. Here is another way to compute 24·25:

   24·25 = 2400/4 = 1200/2 = 600.

   The rule naturally applies to 2-digit numbers as well. 75² = 5625 (since 7·8 = 56).

9. Product of 10a+b and 10a+c where b+c = 10.
   Similar to the squaring of numbers that end with 5:

   (10a + b)(10a + c) = 100a2 + 10a·(b + c) + bc = 100a(a + 1) + bc.

   For example, compute 113×117, where a = 11, b = 3, and c = 7. First compute 11·(11 + 1) = 11·12 = 132 (since 3 = 1 + 2). Next, append 21 (= 3×7) to the right of 132 to get 13221!

   Another example: compute 242×248, with a = 24, b = 2, and c = 8. Then

   24·(24 + 1) = 242 + 24 = 576 + 24 = 600.

   Therefore 242×242²=60016.

10. Product of two one-digit numbers greater than 5.
    This is a rule that helps remember a big part of the multiplication table. Assume you forgot the product 7·9. Do this. First find the excess of each of the multiples over 5: it's 2 for 7 (7 - 5 = 2) and 4 for 9 (9 - 5 = 4). Add them up to get 6 = 2 + 4. Now find the complements of these two numbers to 5: it's 3 for 2 (5 - 2 = 3) and 1 for 4 (5 - 4 = 1). Remember their product 3 = 3·1. Lastly, combine thus obtained two numbers (6 and 3) as 63 = 6·10 + 3.

    The explanation comes from the following formula:

    (5 + a)(5 + b) = 10(a + b) + (5 - a)(5 - b)

    In our example, a = 2 and b = 4.

11. Product of two 2-digit numbers.
    

    1. If the numbers are not too far apart, and their difference is even, one might use the well known formula

       (a + n)(a - n) = a2 - n2.

       a here is the average of the two numbers.
       For example,

       28·24 = 262 - 22 = 676 - 4 = 672

       since 26 = (24 + 28)/2. Also,

       19·31 = 252 - 62 = 625 - 36 = 589

       since 25 = (19 + 31)/2.

    2. If the difference is odd use either n(m + 1) = nm + n or n(m - 1) = nm - n.
       For example,

       7·34 = 37·35 - 37 = 362 - 12 - 37 = 1296 - 1 - 37 = 1258.

       On the other hand,

       37·34 = 37·33 + 37 = 352 - 22 + 37 = 1225 - 4 + 37 = 1258.

       
       

    3. Babilonian method: ab = [(a + b)² - (a - b)²]/4. For example,

       33×32 = [652 - 12]/4 = (4225 - 1)/4 = 4224/4 = 1056.

       
       
    
    

12. Product of numbers that only differ in units.
    If the numbers only differ in units and the sum of the units is 10, like with 53 and 57 or 122 and 128, then think of them as, say 10a + b and 10a + c, where b + c = 10. The product (10a + b)(10a + c) is given by

    (10a + b)(10a + c) = 100a2 + 10a(b + c) + bc = 100a(a + 1) + bc.

    Thus to compute 53 times 57 (a = 5, b = 3, c = 7), multiply 5×(5 + 1) to get 30. Append to the result (30) the product of the units (3·7 = 21) to obtain 3021. Similarly,

    122·128  =  12·13·100 + 2·8 = 15616.

    
    

13. Product of numbers close to 100.
    Say, you have to multiply 94 and 98. Take their differences to 100: 100 - 94 = 6 and 100 - 98 = 2. Note that 94 - 2 = 98 - 6 so that for the next step it is not important which one you use, but you'll need the result: 92. These will be the first two digits of the product. The last two are just 2×6 = 12. Therefore, 94×98 = 9212.

    The same trick works with numbers above 100: 93×102 = 9486. First find the differences: 100 - 93 = 7 and 100 - 102 = -2. Then subtract one of the differences from the other number, e.g. 93 - (-2) = 95. This intends to show the first two digits of the product, i.e., the number 9500. Add to this the product of 7 and -2, or -14: 9500 - 14 = 9486.

    This works because

    (100 - a)(100 - b) = (100 - a)·100 - (100 - a)·b   = (100 - a)·100 - 100b + ab   = (100 - a - b)·100 + ab.

    
    

14. Multiplying by 11.
    To multiply a 2-digit number by 11, take the sum of its digits. If it's a single digit number, just write it between the two digits. If the sum is 10 or more, do not forget to carry 1 over.
    For example, 34·11 = 374 since 3 + 4 = 7. 47·11 = 517 since 4 + 7 = 11.

15. Faster subtraction.
    Subtraction is often faster in two steps instead of one.
    For example,

    427 - 38 = (427 - 27) - (38 - 27) = 400 - 11 = 389.

    A generic advice might be given as "First remove what's easy, next whatever remains". Another example:

    1049 - 187 = 1000 - (187 - 49) = 900 - 38 = 862.

    
    

16. Faster addition.
    Addition is often faster in two steps instead of one.
    For example,

    487 + 38 = (487 + 13) + (38 - 13) = 500 + 25 = 525.

    A generic advice might be given as "First add what's easy, next whatever remains". Another example:

    1049 + 187 = 1100 + (187 - 51) = 1200 + 36 = 1236.

    
    

17. Faster addition, #2.
    It's often faster to add a digit at a time starting with higher digits. For example,

    583 + 645 = 583 + 600 + 40 + 5   = 1183 + 40 + 5   = 1223 + 5   = 1228.

    
    

18. Multipliply, then subtract.
    When multiplying by 9, multiply by 10 instead, and then subtract the other number. For example,

    23·9 = 230 - 23 = 207.

    The same applies to other numbers near those for which multiplication is simplified:

    	23·51	= 23·50 + 23
    		= 2300/2 + 23
    		= 1150 + 23
    		= 1173.
    	87·48	= 87·50 - 87·2
    		= 8700/2 - 160 - 14
    		= 4350 - 160 - 14
    		= 4190 - 14
    		= 4176.

    
    

19. Multiplication by 9, 99, 999, etc.
    There is another way to multiply fast by 9 that has an analogue for multiplication by 99, 999 and all such numbers. Let's start with the multiplication by 9.

    To multiply a one digit number α by 9, first subtract 1 and form β = α - 1. Next, subtract β from 9: γ = 9 - β. Then just write β and γ next to each other:

    9α = βγ.

    For example, find 6×9 (so that α = 6.) First subtract: 5 = 6 - 1. Subract the second time: 4 = 9 - 5. Lastly, form the product 6×9 = 54.

    Next, find 37×99. First, subtract 1: 36 = 37 - 1. Then subtract 63 = 99 - 36. Lastly, form the product: 37×99 = 3663.

    Why does this work? For the multiplication by 9, βγ = 10β + γ:

    βγ = 10β + γ   = 10(α - 1) + (9 - (α - 1))   = 10α - 10 + 10 - α   = 9α,

    as required. Similarly, for a 2-digit α:

    βγ = 100β + γ   = 100(α - 1) + (99 - (α - 1))   = 100α - 100 + 100 - α   = 99α.

    Do try the same derivation for a three digit number. As an example,

    543×999 = 1000×542 + (999 - 542)   = 999×542 + 999   = 999×543

    just by using the distributive law twice.

References

1. A. Benjamin and M. Shermer, Secrets of Mental Math, Three Rivers Press, 2006.
2. A. Benjamin and M. Shermer, Mathemagics, Lowell House, 1993
3. M. Gardner, Mathematical Carnival, Vintage Books, 1977.
4. E. H. Julius, Rapid Math Tricks And Tips, John Wiley & Sons, 1992
5. E. H. Julius, More Rapid Math Tricks And Tips, John Wiley & Sons, 1992
6. S. Flansburg, Math Magic, William Morrow and Co, NY, 1993
7. S. Flansburg, Math Magic for Your Kids, Harper Paperbacks, 1998

On Internet

1. Calculation Tips & Tricks
2. Multiplication Tips & Tricks