how to play magic number??

Road Games
First, ask someone to choose a four-digit numbers (eg 8552). Then, ask him to reduce these numbers with the numbers constituent (example: 8552-8 - 5 - 5-2 = 8532). Furthermore, the results obtained (8532), ask him to hide a number and mention the number remaining. Then make him surprised him by saying the figures hide quickly. (Example: she hides the number 5 of 8532, and mentions three remaining digits: 8, 3, and 2. Then you quickly said, "five figures right ....").
Tricks
How? Are you quite surprised with the "magic" at the top? The trick turned out to be easy ... How? Sum the given number (8 3 2 = 13), then look for number nine nearest multiple of that number but greater than or equal to that number (in this case number and the nearest multiple of 9 which is greater than or equal to 13 is 18). Then, subtract 18 to 13 and you'll get the answer: FIVE!
For the record, you will experience some difficulties if the number from three numbers given are multiples of 9. If that happens, then the numbers are hidden in between two possibilities, 0 or 9. You can outsmart by asking, "major figures right ...?". If the response is negative, meaning the numbers 0 and if the response is positive, meaning the number 9. This is only one way, you can be creative in other ways
ExplanationWell, if you're curious why the above trick work? I will explain. First, you need to know are the characteristics of a number is divisible by 9 is the number of constituent numbers divisible by 9 also. Do not believe? Try it: 81 is divisible by 9, then 8 1 = 9 9 is also divisible. 87 651 is divisible by 9 because 8 7 6 5 1 = 27 is divisible by 9. 8421 is not divisible by 9 because 8 4 2 1 = 15 not divisible by 9.
Now you already know the characteristics of the number is divisible by 9. We return to the game above. Suppose the number chosen is "abcd". This number can be written into 100B 1000a 10c d. Then, subtract numbers with the numbers mean constituent (1000a 100B 10c d) - a - b - c - d = 999a 99b 9c = 9 (111a 11b c). That is, the result must have been a number is divisible by 9, then the number of the figures making up of the results are divisible by 9.