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November 28, 2015
Posted by **maddi 5** on Thursday, October 25, 2012 at 4:38pm.

- math -
**Reiny**, Thursday, October 25, 2012 at 5:48pmI am curious at what level of math this is from.

It looks like you are looking at divisibility of numbers

Here are some basic rules:

- any number of course is divisible by 1

- to be divisible by 2 , the number has to be even

so the 2nd digit must be even

- to be divisible by 3, the sum of the digits of the number has to be divisible by 3

- to be divisible by 4, the last 2 digits must be divisible by 4 , and the last digit must of course be even

- to be divisible by 5, the last digit must be either 0 or 5, but we don't have a 0, so the 5th digit MUST be 5, ahhh, that's a start

- to be divisible by 6, it must be divisible by 2 AND by 3, that is, it must be even and pass the divisible by 3 test.

- to be divisible by 7 ,,,, there is a way, but much too complicated to state in simple terms

- to be divisible by 8, the last 3 digits must be divisible by 8, and of course the last digit must be even

- to be divisible by 9, the sum of the digits must be divisible by 9, which in this case is true no matter how they are arranged.

so far we know:

the 5th digit must be 5

the 2nd, the 4th, the 6th, and the 8th digits must be even

e.g. The 1 cannot be in the place it is in

I suggest you cut out little pieces of paper, label them 1, 2, ... 9 and try it.

Let me know what number you come up with