Secrets of Mental Math
Notes from the book Secrets of Mental Math  Arthur Benjamin (2006)
Some quick tricks
Multiplying by 11:

45x11 = (first digit) (sum of the digits) (last digit) = (4) (4+5) (5) = 495
If the sum exceeds 10, carry the 1 to the left:
 67x11 = (6) (6+7) (7) = (6) (13) (7) = (6+1) (3) (7) = 737

Works for larger products as well.
 453x11 = 4 (4+5) (5+3) 3 = 4983
 681x11 = 6 (14) (9) 1 = 7491
Multiplying two digit numbers
 If two numbers start with the same digit, and their last digits add up to 10, first multiply the first digit (x) and x+1, and then multiply their last digits, and put them together:
 37x33 = (3x4) (7x3) = 1221
 As a special case, this is very helpful when squaring a 2digit number ending in 5
 65x65 = (6x7) (25) = 4225
Calculating tips
 To quickly calculate 15% of, say 42:
 10% of 42 = 4.2
 5% = 4.2/2 = 2.1
 15% = 4.2 + 2.1 = 6.3
Mathemagical tricks
Psychic math
Ask a person to think of a number (preferably onedigit or twodigit) and ask them to:
 double it,
 add 12,
 divide the total by 2,
 subtract the original number.
The sequence will always produce the number 6.
The sequence goes like this: x > 2x > 2x+12 > x+6 > x+6x > 6
Magic 1089
 Write down a threedigit number where the digits are decreasing (like 851 or 973),
 Reverse that number and subtract it from the first number,
 Take that answer and add it to the reverse of itself.
No matter what the original number is, the sequence will always produce 1089!
Step 1 and 2 are alebraically equivalent to: 100a+10b+c  (100c+10b+a) = 99(ac) which must be one of the following multiples of 99: 198, 297, 396, 495, 594, 693, 792, or 891, each one of which will produce 1089 after adding it to the reverse of itself.
Missing digit tricks
Many variations of missing digit tricks can be performed by using the fact that a number is a multiple of 9 if and only if its digits sum to a multiple of 9.
For this, you start by forcing a person to come up with a large multiple of 9, asking them to tell you all the digits except one in random order, adding those and subtracting it from the nearest multiple of 9.
For example, ask them to multiply any threedigit number by 1089 (a multiple of 9) from the trick mentioned above. (Say they secretly multiply 1089x573 = 623,997). Ask them to choose a secret digit and call out the rest in any order. (Say they call out, "2...9...3...7...9") Add those to produce 2+9+3+7+9 = 30. The nearest multiple of 9 is 36 which is 6 more than 30. The missing digit must be 6.
If the digits called out add up to a multiple of 9, then there is no way of determining whether the missing digit is 0 or 9. In this case, you merely say, "You didn’t leave out a zero, did you?" If they say yes, you're done. If not, you know the missing digit is nine.
Some subtle ways to force the volunteer to end up with a multiple of 9:
 randomly choose a sixdigit number, scramble its digits, then subtract the smaller sixdigit number from the larger one.
 secretly choose a fourdigit number, reverse the digits, then subtract the smaller number from the larger.
 multiply onedigit numbers randomly until the product is seven digits long. This is not “guaranteed” to produce a multiple of 9, but in practice it will do so at least 90% of the time (the chances are high that the onedigit numbers being multiplied include a 9 or two 3s or two 6s, or a 3 and a 6).
Rapid cube root of twodigit numbers
This trick relies on the fact that the cubes of single digit whole numbers end in unique digits.
Example: Cube root of 474552 (AAABBC) = 78
Steps:
 Cube just lower than AAA (474) = 343 = 7^3
 Cube ending in C (2) = 512 = 8^3
Quick square root of twodigit numbers
Example: Square root of 7921 = 89
 Look at the numbers preceding the last two digits (79).
 79 lies between 8x8=64 and 9x9=81, we know the answer lies in the 80s. There are two numbers whose squares ends in 1: 1 and 9. So the answer must be 81 or 89.
 Compare the original number with the square of 85 (80x90 + 25 = 7225). Since 7921 > 7225, the answer must be 89.
An amazing sum
Start with a prediction 2247 and three sets of nine numbers each:
 A: 4286 5771 9083 6518 2396 6860 2909 5546 8174
 B: 5792 6881 7547 3299 7187 6557 7097 5288 6548
 C: 2708 5435 6812 7343 1286 5237 6470 8234 5129
 Choose a random number from each set (say 4286, 5792, and 5435).
 Take one digit from each fourdigit number in order (say 895).
 Repeat this process three more times with the remaining three numbers, resulting in three more threedigit numbers such as 453, 224, and 675.
 Add the four numbers to produce 2247, the original prediction.
A B C 8 9 5 4 5 3 2 2 4 + 6 7 5  2 2 4 7
This trick works because each set of numbers sums to the same total (A > 20, B > 23, C > 17). With person C’s numbers in the right column totaling to 17, you will always put down the 7 and carry the 1. With person B’s numbers totaling to 23, plus the 1, you will always put down the 4 and carry the 2. Finally with person A’s numbers totaling to 20, adding the 2 gives you a total of 2247!