All three of the following magical tricks deal with the idea of digital roots...and Casting Out 9's. If you have not played with last week's problem, please do so first.
Trick 1: The teacher asks a student to pick a 3-digit number and tell her the number as well. Now, she asks them to divide their number by 9 but not tell her the answer, especially the remainder. But, she can instantly tell them the remainder! How?
Also, try this trick with n-digit numbers...does it work?
Trick 2: On a piece of paper, a teacher writes the number 0 and places it in her pocket. Then, she asks a student to first write down a 4-digit number then randomly mix the digits any way desired to produce a second 4-digit number. She then has the student subtract the smaller number from the larger number and divide the result by 9. Then, magically she pulls the piece of paper from her pocket to reveal that the remainder was 0. How did she do this?
Does the trick work with n-digit numbers?
Trick 3: I have used this trick many times with success! The teacher asks a student to write down a 4-digit number, again mix the digits to form a new 4-digit number, subtract the larger number from the smaller, cross out any digit in the difference, and then state all the other digits in the difference (in any order). From those digits, the teacher can magically state the crossed out digit. How?
Why is it a problem if the crossed out digit is 9 or 0?
Does the trick work for n-digit numbers?
Hint: Try each trick out...many times. Look for patterns...and have "Casting out 9's" or "digital roots" on your brain.
Trick 1: You merely add up the three digits, cast out nines (i.e. find the digital root), and that is the remainder.
Trick 2: Briefly, the two numbers would have had the same remainder if divided by 9, thus the difference of the numbers also is divisible by 9 (something called the distributive property?). Suppose the original two numbers were 9x+n and 9y+n, where n is the common remainder or digital root. Then, the difference of the two numbers is (9x+n) - (9y+n) = 9x-9y = 9(x-y) + 0, or has a digital root of 0.
Trick 3: Add up the stated digits, find their digital root...and then the missing digit is the digit needed to make 9 (or 0). That is, if the digits were 3, 5, 7, their sum is 15 with a digital root of 6, so the crossed out digits must have been a 3, as 3+6 = 9.