# Number 9 phenomenon

I have seen this number "trick" throughout the years and have never been able to figure out the mechanism. I'm sure most of you are familiar with it, and probably have a simple solution. Here it is for those who are not aware:

take any number greater than 9

add up all of the digits of this number

subtract the sum from the original number

simplify this number by adding the digits until you get down to a single digit...it will ALWAYS be 9

example:

384773

3+8+4+7+7+3 = 32

384773-32 = 384741

3+8+4+7+4+1 = 27

2+7 = 9

Anyone know what's going on here?

Last edited:

CRGreathouse
Homework Helper
Anyone know what's going on here?

10 = 9*1 + 1, so x + 10 is divisible by 9 exactly when x + 1 is divisible by 9. This lets you move the tens place to the ones place:

37 = 27 + 10 -> 27 + 1 = 18 + 10 -> 18 + 1 = 9 + 10 -> 10 = 0 + 10 -> 1

so 37 is not divisible by 9 (it leaves a remainder of 1). In fact, this lets you move the hundreds place down to the 1s place in the same way, since 100 = 9*11 + 1. Etc.

For base b, this trick works for divisibility by b-1. Since you're using base 10, it works for 9. If you use hexadecimal, it works for divisibility by 15.

10 = 9*1 + 1, so x + 10 is divisible by 9 exactly when x + 1 is divisible by 9. This lets you move the tens place to the ones place:

37 = 27 + 10 -> 27 + 1 = 18 + 10 -> 18 + 1 = 9 + 10 -> 10 = 0 + 10 -> 1

so 37 is not divisible by 9 (it leaves a remainder of 1). In fact, this lets you move the hundreds place down to the 1s place in the same way, since 100 = 9*11 + 1. Etc.

For base b, this trick works for divisibility by b-1. Since you're using base 10, it works for 9. If you use hexadecimal, it works for divisibility by 15.

So it's the system of mathematics that we're using? Not a naturally occurring anomaly?

CRGreathouse
Homework Helper
So it's the system of mathematics that we're using? Not a naturally occurring anomaly?

Anything that has to do with the decimal digits of a number will naturally need to use the fact that the base is 10.

You could also look at it this way:
1) Assume you have a 4 digit number ABCD
2) This can be represented by 1000A + 100B + 10C + D
3) If you add the digits together, you get A + B + C + D
4) Subtract the result of 3) from the result of 2). You get: (1000A + 100B + 10C + D) - (A + B + C + D) = 999A + 99B + 9C
5) Which is divisible by 9: 9 x (99A + 9B + C)
6) Any number evenly divisible by 9 has the following property: add all of the digits of the number to get a new number, continue this until you only have one digit, the result is always 9

(Now the job is to prove #6)

You can easily see that this will work for any integer with 'n' digits such that n > 1

Integral
Staff Emeritus