Question

For any given even number `sqrtA` that has `m` digits and `A` with `n` digits, is there a relationship between `m` and `n`?

Answer 1

• if `n` is even, `m=n/2`
• if `n` is odd, `m=(n+1)/2`

where `m` is the number of digits in `sqrtA`

Explanation:

Let's first look at a few terms and see if we can spot a pattern (I'm including the variable `m` which is the number of digits in `sqrtA`):

`((A, n,sqrtA, m),(4,1,2,1),(16,2,4,1),(36,2,6,1),(64,2,8,1),(100,3,10,2),(144,3,12,2),(vdots, vdots, vdots, vdots))`

Since `A` is even, then `sqrtA` will also be even. And so what I'll do next is look at what values of `A` we generate for different values of `sqrtA` with different `m`:

`((A, n,sqrtA, m),(4,1,2,1),(16,2,4,1),(vdots, vdots, vdots, vdots),(64,2,8,1),(100,3,10,2),(vdots, vdots, vdots, vdots),(900,3,30,2),(1024,4,32,2),(vdots, vdots, vdots, vdots),(9604,4,98,2),(10000,5,100,3))`

And so we can see relationships:

`n=1, m=1`
`n=2, m=1`
`n=3, m=2`
`n=4, m=2`
`n=5, m=3`

and so on.

And so we can calculate `m` for any `n`, where,

• if `n` is even, `m=n/2`
• if `n` is odd, `m=(n+1)/2`

Let's test it out:. Let's say `sqrtA=45678, m=5=>A=2086479684, n=10 color(green)sqrt`