Multiples of Nine, Digit Sum ```Name: Sharvind Status: student Grade: 9-12 Country: Malaysia Date: N/A ``` Question: why when a number when times by 9,the numbers in the answers when added until a single digit is found,the number is 9?..eg. 538 x 9 = 4842,4+8+4+2=18,1+8=9...and why its only for the number 9 and not for other numbers? Replies: Sharvind, Think of the number 531, for example. You can write 531 as 500 + 30 + 1 = 5X100 + 3X10 + 1. Since 100 = 99 + 1 and 10 = 9 + 1, 531 can be written as (5X99 + 5X9) + (3X9 + 3X1) + 1X1 = (5X99 + 3X9) + (5+3+1). Since 9 divides evenly into 99 and 9, it surely goes into the 5X99 + 5X9 part. So 9 goes into the entire number if and only if it goes into the (5+3+1) part. This works for any whole number expressed in decimal notation. For example, 8347 = (8X999 + 3X99 + 4X9) + (8 + 3 + 4 + 7). Since 9 goes into the first part but not into the second (the digit sum), it cannot go into the entire number. By the way, the same rule works for divisibility by 3. For example, with the number 8347, 3 goes into the (8X999 + 3X99 + 4X9) part, and therefore, 3 must go into 8347, because it goes into the (8+3+4+7) part. You can use other bases too. For example, in base 5, the same trick would work for 4 and 2. For example, the number 1020304 is a base 5 number with digit sum 1+2+3+4=20 (base 5). This would expanded as (all numbers in base 5) 1020304=(4+3+2+1)+3X(44)+2X4444+1X444444. Since the second part is clearly divisible by 4 ( and 2), the number is divisible by 4 or 2 if and only if (4+3+2+1) is. (You need to be careful working with bases that are not prime.) If you iterate the process until a single digit is obtained, then the original number is divisible by 9 if and only if the single digit is 9 (or 0). For example 8347 --> 24 --> 6. So, 8347 is divisible by 9 if and only if 24 is and 24 is divisible by 9 if only 6 is. In the base 5 example, 1020304 is divisible by 4 (or 2) if and only if 20 is. And 20 is divisible by 4 (or 2) if and only if 2 is. (1020304 --> 20 --> 2) So 1020304 is divisible by 2 but not 4. Regards, Paul Beem Sharvind, Every integer product of nine has single digits that add up to a product of nine, which then repeats again until you have the single digit nine, because of how carrying works. When you carry, you essentially subtract ten from a column and add one to the next column. This is like subtracting nine from the total. When carrying happens because you have added nine, then really the total of digits hasn't changed. For any other single digit, the digit total changes when you carry. When adding nine, the digit total only changes when the last digit is a zero. In that case, the digit total increases by nine, but the digit total is still a multiple of nine. When nine is added to a number with a digit total of nine, the digit total stays a multiple of nine even when you have to carry. Starting with nine, which does have a digit total of nine, you can thus add nine to it as many times as you like and keep the digit total a multiple of nine. Doing this over and over will take you through every multiple of nine. This multiple, which might be 27, is a smaller multiple of nine. Its digits can then be reduced to an even smaller multiple of nine, eventually reaching nine itself. Dr. Ken Mellendorf Physics Instructor Illinois Central College It has to do with remainder 9. I offer you some interesting counter examples: 537 x 9 = 4833 = 4 + 8 + 3 + 3 = 18 = 1 + 8 = 9 536 x 9 = 4824 = 4 + 8 + 2 + 4 = 18 = 1 + 8 = 9 539 x 9 = 4851 = 4 + 8 + 5 + 1 = 18 = 1 + 8 = 9 Also notice that for any number that is a multiple of '9', the sum of the last two digits is '6'. The sum of the last three digits is 14. The sum of all four digits is 18. It has to do with a concept of modular arithmetic. Your math teacher may be able to dig into this with you. Vince Calder Click here to return to the Mathematics Archives

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