Are my findings valid? If so they must be on some maths logic that is beyond me...
So I was quizzing my kids (7&9) on seeing if a number was divisible by 3. They then asked me about 4 but the standard 'trick' (last 2 digits are divisible by 4) is not 'easy' for them so I observed the following:
On any length number:
$F$ = Penultimate digit
$V$ = last digit
If $(F + \frac{V}{2})$ is EVEN,
Then the original number is divisible by $4$.
Eg
$36 = 3 + 6/2$ which is even
$46 = 4 + 6/2$ which is odd
Intrigued I tested this for $8$
$F =$ antepenultinate digit
$V =$ last $2$ digits
If $(F + \frac{V}{2*2})$ is Even
Then the original number is divisible by $8$
Eg
$816 = 8 + \frac{16}{4}$ = even
$820 = 8 + \frac{20}{4}$ = odd
I tested again for $16$ and the pattern continued,
So
$2^x$ Is a factor of INTEGER if...
$F = INTEGER [INTEGER.Length-x+1]$ (single digit)
$V =\text{last }(x - 1) \text{ digit(s)}$
If $F + \frac{V}{2^x}$ = EVEN
Then
$2^x$ is a factor of INTEGER
Was I right? Is there a proof?
What logic is this based on - simple times tables?