# Number question

1. Sep 10, 2007

### TR345

when you take a number n=2,3,4,5,6,7,8,or 9 and multiply by 2 then mulitply the product again by two and so on. Then add the individual digits. You get interesting patterns for each number 2-9. well 2 and 4 and 8 are the same yet have no obvious patern. 7 is unique, 3 and 6 are the same and create an interesting sequence, but when it comes to 9, the sum is always a multiple of 9. 9-9,18-9,36-9,72-9, 144-9,288-18,576-18,1152-9,2304-9,4608-18 .............

If the process is repeated, will there ever be a number where the individual digits don't equal a multiple of 9.

7-7
14-5
28-10
56-11
112-4
224-8
448-16
896-23

2. Sep 10, 2007

### TR345

Also if you do the same thing except multiply by 3. Then the patern created when n=7 is that all the sloutions are multiples of 9.

3. Sep 10, 2007

### ramsey2879

except for 21 which is a multiple of 3. In base 10 if N = ..DCBA = A + 10B +100C + 1000D .... But N mod 3 or 9 is the same as A + B + C + D since you subtracted 9B + 99C + 999D. Since your N is a multiple of 3 or nine and since you subtracted a multiple of 9 you must end up with a multiple of 3 or 9 respectively.

Last edited: Sep 10, 2007
4. Sep 10, 2007

### ramsey2879

The pattern is even more interesting if you keep adding the digits until you get a single digit .e.g. 448 -> 16 -> 7. You will find that since 6 numbers less than 9 are not divisible by 9 (that is phi(9) = 6 ) then for the 7th term in your sequence, the sum of the digits will end up being the digit you started with even if you started with a 3, 6 or 9 which don't appear in the sequence unless you start with a 3 or 6, 6 or 3, or 9 respectively.
As I noted before it all has to do with the relation between base 10 and mod 9.
Edit: Also if you use other multipliers less than 9 (but not itself a multiple of 3) you will get a different sequences that will always equal the number you started with at the 7th term. See Fermat's Little Theorem.

Last edited: Sep 10, 2007