MOG 2017 2

Screenshot 2020-08-08 at 20.36.22

The first thing to notice is that a jiggly number must be even \therefore the only digits we can have are 0,2,4,6,8. We can also disregard 0 as a digit because we can’t have a 0 at the front of an integer.
We can take the even property one step further and say that a jiggly number must be a multiple of 4 (because it is a multiple of 12).
That is it must end in 24,28,44,48,64,68,84,88
Due to the rearrangement property we must be able to take 24,28,...,88 swap the digits but still have a multiple of 4. Otherwise we could just swap the last two digits of our ‘jiggly’ number and we wouldn’t have a multiple of 12.
For example 4224=12\times352 but 4242 is not a multiple of 12.
A quick check will see that this is only true for 44,48,84, and, 88.
So our jiggly number only contains the digits 4 and/or 8
As a multiple of 12, our jiggly number is a multiple of 3, so the sum of its digits is a multiple of 3.
So we can see that all jiggly numbers are a perumutation of 4488
There are 6 of these: (4488,4848,8448,8484,8844,4884)
\therefore only 6 jiggly numbers exist



Really great solution, I did notice a small typo. In the example I think it should say
4224 = 12 x 352

Still a really helpful solution to read :+1:


Thank you!