 Quandaries and Queries I understand that when you pick a number greater than 1 and less than 10; multiply it by 7 and add 23, then add the digits of that number until you get a one digit number. Then multiply that number by 9, add the digits of that number until you get a one digit number, subtract 3 from that number and divide the difference by 3; that this process will always give you the result of 2. Does this have a name or theory for it as to why the answer will always be 2? Hi, The property that makes this work is sometimes called "casting out nines". The property is this: Take any positive integer n, divide it by nine and record the remainder. Call it r1. Take the same integer n and add its digits. Divide the sum of the digits by 9 and record the remainder. call it r2. The conclusion is that r1 = r2. For example if n = 7312 then n divided by 9 is 812 with a remainder of 4, that is r1 = 4. Adding the digits of n gives 7+3+1+2=13. 13 divided by 9 is 1 with a remainder of 4, that is r2 = 4. This r1 = r2 = 4. So, how does this relate to your problem? The first sentence is really unimportant, just do anything that results in a one digit number. Multiply this number by 9 and you have a number that is divisible by 9. Thus if you divide this number by 9 and record the remainder, the remainder is 0. Add the digits of this number. By the property above, the sum of the digits is also divisible by 9. Since this sum is a one digit number it must be 9. Now when you subtract 3 you get 6, then divide by 3 you get 2. Penny Go to Math Central