Multiples of Nine, Digit Sum
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?
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.
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
Illinois Central College
It has to do with remainder 9. I offer you some interesting counter
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.
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Update: June 2012